Square Root Questions

Square and square root practice test:.

The largest of four digit numbers which is a perfect square is (a) 9801 (b) 9904 (c) 9804 (d) 9809 View Answer Ans. (a)

Evaluate √6084 by factorization method . (a) 78 (b) 76 (c) 82 (d) 64 View Answer Ans. (a)

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What is the square root of 8? (a) 2.828 (b) 2.928 (c) 2.728 (d) 2.628 View Answer Ans. (a)

Find the square root of each of 291600 by the long division method: (a) 520 (b) 620 (c) 580 (d) 540 View Answer Ans. (d)

What is the square root of 0.0009? (a) 0.81 (b) 0.27 (c) 0.03 (d) 0.003 View Answer Ans. (c)

Find the 169 square root. (a) 21 (b) 11 (c) 17 (d) 13 View Answer Ans. (d)

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Using the square root ability, find the value 6.328 (a) 2.512 (b) 2.518 (c) 2.517 (d) 2.514 View Answer Ans. (d)

What is the square root of 85? (a) 9.519 (b) 9.419 (c) 9.319 (d) 9.219 View Answer Ans. (d)

Evaluate: √0.9 up to 3 places of decimal. (a) 0.948 (b) 0.958 (c) 0.938 (d) 0.978 View Answer Ans. (a)

A general arranges his soldiers in a row to form a perfect square. He finds that in doing so, 60 soldiers are left out. If the total number of soldiers is 8160, find the number of soldiers in each row. (a) 96 soldiers (b) 92 soldiers (c) 90 soldiers (d) 99 soldiers View Answer Ans. (c)

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What is the square root of 61? (a) 7.910 (b) 7.810 (c) 7.710 (d) 7.610 View Answer Ans. (b)

The area of a square plot is 2304 m 2 . Find the side of the square plot. (a) 44 m (b) 46 m (c) 48 m (d) 42 m View Answer Ans. (c)

Find the square root of 1471369. (a) 1123 (b) 1213 (c) 1323 (d) 1133 View Answer Ans. (b)

What is the square root of 5? (a) 2.536 (b) 2.436 (c) 2.336 (d) 2.236 View Answer Ans. (d)

A school collected $2304 in fees from its students. If each student paid as much fee as there were students in the school, how many students were there in the school? (a) 48 (b) 46 (c) 44 (d) 42 View Answer Ans. (a)

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What is the square root of 100 by prime factorization? (a) 5 (b) 10 (c) 15 (d) 20 View Answer Ans. (b)

Find the smallest number by which 28812 must be divided so that the quotient becomes a perfect square. (a) 9 (b) 7 (c) 6 (d) 3 View Answer Ans. (d)

Find the value of √3 up to three places of decimal. (a) 1.369 (b) 1.123 (c) 1.634 (d) 1.732 View Answer Ans. (d)

Find the square root of 625 by prime factorization. (a) 31 (b) 19 (c) 23 (d) 25 View Answer Ans. (d)

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What is the square root of 162? (a) 12.727 (b) 12.827 (c) 12.927 (d) 12.627 View Answer Ans. (a)

A gardener plants an orchard with 5776 trees. In each row there were as many trees as the number of rows. Find the number of rows. (a) 76 (b) 96 (c) 66 (d) 186 View Answer Ans. (a)

Find the value of √ 0.289 / 0.00121. (a) 170 / 11 (b) 150 / 11 (c) 160 / 11 (d) 140 / 11 View Answer Ans. (a)

Find the square of 32: (a) 864 (b) 954 (c) 1056 (d) 1024 View Answer Ans. (d)

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Find the square root of 729 by prime factorization? (a) 27 (b) 29 (c) 31 (d) 33 View Answer Ans. (a)

What is the square root of 80 upto two decimal points? (a) 8.94 (b) 8.84 (c) 8.74 (d) 8.64 View Answer Ans. (a)

Find the least square number which is exactly divisible by 10, 12, 15 and 18. (a) 750 (b) 800 (c) 850 (d) 900 View Answer Ans. (d)

Find the greatest number of two digits, which is a perfect square. (a) 64 (b) 100 (c) 99 (d) 81 View Answer Ans. (d)

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Find the greatest number of 5 digits, which is a perfect square. (a) 99876 (b) 99866 (c) 99856 (d) 99846 View Answer Ans. (c)

What is the square root of 108 (a) 10.592 (b) 10.492 (c) 10.392 (d) 10.292 View Answer Ans. (c)

Find the greatest number of five digits, which is a perfect square. (a) 99856. (b) 99876 (c) 94766 (d) 99248 View Answer Ans. (a)

Some people contributed $1089. Each person gave as many rupees as they were in number. Find their number. (a) 33 (b) 66 (c) 45 (d) 23 View Answer Ans. (a)

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Find the least number of 4 digits, which is a perfect square. (a) 1054 (b) 1044 (c) 1034 (d) 1024 View Answer Ans. (d)

Find square root of 128. (a) 11.313 (b) 11.413 (c) 11.513 (d) 11.613 View Answer Ans. (a)

Find the smallest number that must be added to 1780 to make it a perfect square. (a) 79 (b) 69 (c) 89 (d) 49 View Answer Ans. (b)

There are 2401 students in a school. P.T. teacher wants them to stand in rows and columns such that the number of rows is equal to the number of columns. Find the number of rows. (a) 33 (b) 64 (c) 49 (d) 51 View Answer Ans. (c)

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What is the square root of 361? (a) 19 (b) 21 (c) 23 (d) 27 View Answer Ans. (a)

Find the square of 35 (a) 1265 (b) 1225 (c) 1355 (d) 1335 View Answer Ans. (b)

Find the cube root of 2744. (a) 24 (b) 34 (c) 14 (d) 16 View Answer Ans. (c)

Find the least square number which is exactly divisible by each one of the numbers 15, 18 and 12. (a) 1800 (b) 3600 (c) 2400 (d) 4200 View Answer Ans. (b)

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Find the square root of 41? (a) 6.403 (b) 6.405 (c) 6.407 (d) 6.409 View Answer Ans. (a)

What is the square root of 28? (a) 5.491 (b) 5.391 (c) 5.291 (d) 5.191 View Answer Ans. (c)

Without adding, find the sum (1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17). (a) 81 (b) 79 (c) 77 (d) 75 View Answer Ans. (a)

Related: general science Biology quiz

Find square root of 41 (a) 6.603 (b) 6.503 (c) 6.403 (d) 6.303 View Answer Ans. (c)

Find the least number of three digits, which is a perfect square. (a) 222 (b) 128 (c) 100 (d) 102 View Answer Ans. (c)

What is the square root of 34 upto two decimal points? (a) 5.53 (b) 5.63 (c) 5.73 (d) 5.83 View Answer Ans. (d)

Related: Puzzle questions

Find the square of 86 (a) 7856 (b) 7786 (c) 7396 (d) 7464 View Answer Ans. (c)

Find square root of 52? (a) 7.211 (b) 7.311 (c) 7.411 (d) 7.511 View Answer Ans. (a)

Find the greatest number of 5 digits, which is a perfect square. 21. (a) 99886 (b) 99876 (c) 99866 (d) 99856 View Answer Ans. (d)

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What is the square root of 29? (a) 5.585 (b) 5.485 (c) 5.385 (d) 5.285 View Answer Ans. (c)

Find the least number of six digits, which is a perfect square. (a) 100499 (b) 100489 (c) 100479 (d) 100469 View Answer Ans. (b)

Square root of 21 ? (a) 1.532 (b) 1.632 (c) 1.732 (d) 1.832 View Answer Ans. (c)

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The area of a square playground in 256.6404 square metres. Find the length of one side of the playground. (a) 16.02 metres (b) 17.02 metres (c) 18.02 metres (d) 19.02 metres View Answer Ans. (a)

What is the square root of 14? (a) 3.441 (b) 3.541 (c) 3.641 (d) 3.741 View Answer Ans. (d)

Which of the following numbers are squares of even numbers? (a) 324 (b) 6561 (c) 4489 (d) 373758 Ans: (a)

Related: Ratio proportion formula questions

What will be the square root of 26? (a) 5.099 (b) 5.199 (c) 5.299 (d) 5.399 View Answer Ans. (a)

Express the following as the sum of two consecutive integers: 19 2 (a) 180 + 181 (b) 60 + 61 (c) 84 + 85 (d) 220 + 221 View Answer Ans. (a)

Find the least must be added to 7900 to obtain a perfect square. (a) 67 (b) 87 (c) 97 (d) 77 View Answer Ans. (d)

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Square root of negative 1? (a) i (b) 1 (c) 0 (d) -1 View Answer Ans. (a)

Find the square root of 7 correct upto two decimal places. (a) 2.65 (b) 2.75 (c) 2.85 (d) 2.95 View Answer Ans. (a)

Find the square root of 60? (a) 7.945 (b) 7.845 (c) 7.745 (d) 7.645 View Answer Ans. (c)

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A welfare association collected $202500 as a donation from the residents. If each paid as many rupees as there were residents, find the number of residents. (a) 480 (b) 470 (c) 460 (d) 450 View Answer Ans. (d)

Find the square root 2. (a) 1.314 (b) 1.414 (c) 1.514 (d) 1.614 View Answer Ans. (b)

Find the square roots of 9216 by the Prime factorisation method. (a) 96 (b) 106 (c) 116 (d) 126 View Answer Ans. (a)

Related: Reasoning questions

Find square root of 324. (a) 18 (b) 16 (c) 14 (d) 22 View Answer Ans. (a)

What is the square root of 96? (a) 9.99 (b) 9.89 (c) 9.79 (d) 9.69 View Answer Ans. (c)

Find the smallest number by which 180 must be multiplied so that it becomes a perfect square. Also find the square root of the perfect square so obtained. (a) 50 (b) 40 (c) 30 (d) 20 View Answer Ans. (c)

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Square root of 35? (a) 5.916 (b) 5.816 (c) 5.716 (d) 5.616 View Answer Ans. (a)

Find square root of 89. (a) 9.333 (b) 9.433 (c) 9.533 (d) 9.633 View Answer Ans. (b)

Find the smallest number by which 396 must be multiplied so that the product becomes a perfect square. (a) 13 (b) 19 (c) 16 (d) 11 View Answer Ans. (d)

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What is the square root of 64 by prime factorization? (a) 9.4 (b) 8 (c) 8.4 (d) 9.4 View Answer Ans. (b)

Find the square root of 98. (a) 9.899 (b) 9.799 (c) 9.699 (d) 9.599 View Answer Ans. (a)

A PT teacher wants to arrange maximum possible number of 6000 students in a field such that the number of rows is equal to the number of columns. Find the number of rows if 71 were left out after arrangement. (a) 77 rows (b) 73 rows (c) 71 rows (d) 69 rows View Answer Ans. (a)

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Find square root of 1000 by prime factorization. (a) 100 (b) 25 (c) 31.622 (d) 36.422 View Answer Ans. (c)

Find the square root of 58 upto three decimal points? (a) 7.615 (b) 7.715 (c) 7.815 (d) 7.915 View Answer Ans. (a)

Find the square roots of 169 by the method of repeated subtraction. (a) 13 (b) 17 (c) 19 (d) 21 View Answer Ans. (a)

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225 square root? (a) 13 (b) 17 (c) 15 (d) 19 View Answer Ans. (c)

What square root of 37 upto two decimal points? (a) 6.08 (b) 6.18 (c) 6.19 (d) 6.09 View Answer Ans. (a)

Find square root of 3 up to three decimal places. (a) 1.739 (b) 1.832 (c) 1.933 (d) 1.732 View Answer Ans. (d)

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What is the square root of 81? (a) 13 (b) 11 (c) 7 (d) 9 View Answer Ans. (d)

Find the square root of 180. (a) 13.416 (b) 14.416 (c) 15.416 (d) 16.416 View Answer Ans. (a)

What is the square root of 68? (a) 8.546 (b) 8.446 (c) 8.346 (d) 8.246 View Answer Ans. (d)

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Find the square root of 576. (a) 24 (b) 25 (c) 26 (d) 27 View Answer Ans. (a)

Square Root Chart:

square root infographic

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How to Solve Square Root Problems

Last Updated: January 2, 2024 Fact Checked

This article was co-authored by David Jia . David Jia is an Academic Tutor and the Founder of LA Math Tutoring, a private tutoring company based in Los Angeles, California. With over 10 years of teaching experience, David works with students of all ages and grades in various subjects, as well as college admissions counseling and test preparation for the SAT, ACT, ISEE, and more. After attaining a perfect 800 math score and a 690 English score on the SAT, David was awarded the Dickinson Scholarship from the University of Miami, where he graduated with a Bachelor’s degree in Business Administration. Additionally, David has worked as an instructor for online videos for textbook companies such as Larson Texts, Big Ideas Learning, and Big Ideas Math. There are 11 references cited in this article, which can be found at the bottom of the page. This article has been fact-checked, ensuring the accuracy of any cited facts and confirming the authority of its sources. This article has been viewed 384,336 times.

While the intimidating sight of a square root symbol may make the mathematically-challenged cringe, square root problems are not as hard to solve as they may first seem. Simple square root problems can often be solved as easily as basic multiplication and division problems. More complex square root problems, on the other hand, can require some work, but with the right approach, even these can be easy. Start practicing square root problems today to learn this radical new math skill!

Understanding Squares and Square Roots

David Jia

  • Try squaring a few more numbers on your own to test this concept out. Remember, squaring a number is just multiplying it by itself. You can even do this for negative numbers. If you do, the answer will always be positive. For example, (-8) 2 = -8 × -8 = 64 .

Step 2 For square roots, find the

  • As another example, let's find the square root of 25 (√(25)). This means we want to find the number that squares to make 25. Since 5 2 = 5 × 5 = 25, we can say that √(25) = 5 .
  • You can also think of this as "undoing" a square. For example, if we want to find √(64), the square root of 64, let's start by thinking of 64 as 8 2 . Since a square root symbol basically "cancels out" a square, we can say that √(64) = √(8 2 ) = 8 .

Step 3 Know the difference between perfect and imperfect squares.

  • On the other hand, numbers that don't give whole numbers when you take their square roots are called imperfect squares . When you take one of these numbers' square roots, you usually get a decimal or fraction. Sometimes, the decimals involved can be quite messy. For instance, √(13) = 3.605551275464...

Step 4 Memorize the first 10-12 perfect squares.

  • 1 2 = 1 × 1 = 1
  • 2 2 = 2 × 2 = 4
  • 3 2 = 3 × 3 = 9
  • 4 2 = 4 × 4 = 16
  • 5 2 = 5 × 5 = 25
  • 6 2 = 6 × 6 = 36
  • 7 2 = 7 × 7 = 49
  • 8 2 = 8 × 8 = 64
  • 9 2 = 9 × 9 = 81
  • 10 2 = 10 × 10 = 100
  • 11 2 = 11 × 11 = 121
  • 12 2 = 12 × 12 = 144

Step 5 Simplify square roots by removing perfect squares when possible.

  • Let's say that we want to find the square root of 900. At first glance, this looks very difficult! However, it's not hard if we separate 900 into its factors. Factors are the numbers that can multiply together to make another number. For instance, since you can make 6 by multiplying 1 × 6 and 2 × 3, the factors of 6 are 1, 2, 3, and 6.
  • Instead of working with the number 900, which is somewhat awkward, let's instead write 900 as 9 × 100. Now, since 9, which is a perfect square, is separated from 100, we can take its square root on its own. √(9 × 100) = √(9) × √(100) = 3 × √(100). In other words, √(900) = 3√(100) .
  • We can even simplify this two steps further by dividing 100 into the factors 25 and 4. √(100) = √(25 × 4) = √(25) × √(4) = 5 × 2 = 10. So, we can say that √(900) = 3(10) = 30 .

Step 6 Use imaginary numbers for the square roots of negative numbers.

  • Note that although imaginary numbers can't be represented with ordinary digits, they can still be treated like ordinary numbers in many ways. For instance, the square roots of negative numbers can be squared to give those negative numbers, just like any other square root. For example, i 2 = -1

Using Long Division-Style Algorithms

Step 1 Arrange your square root problem like a long division problem.

  • Start by writing out your square root problem in the same from as a long division problem. For example, let's say that we want to find the square root of 6.45, which is definitely not a convenient perfect square. First, we'd write an ordinary radical symbol (√), then we'd write our number underneath it. Next, we'd make a line above our number so that it's in a little "box" — just like in long division. When we're done, we should have a long-tailed "√" symbol with 6.45 written under it.
  • We'll be writing numbers above our problem, so be sure to leave space.

Step 2 Group digits into pairs.

  • In our example, we would divide 6.45 into pairs like this: 6-.45-00 . Note that there is a "leftover" digit on the left — this is OK.

Step 3 Find the biggest number whose square is less than or equal to the first

  • In our example, the first group in 6-.45-00 is 6. The biggest number that is less than or equal to 6 when squared is 2 — 2 2 = 4. Write a "2" above the 6 under the radical.

Step 4 Double the number you just wrote down, then drop it down and subtract it.

  • In our example, we would start by taking the double of 2, the first digit of our answer. 2 × 2 = 4. Next, we would subtract 4 from 6 (our first "group"), getting 2 as our answer. Next, we would drop down the next group (45) to get 245. Finally, we would write 4 once more to the left, leaving a small space to add onto the end, like this: 4_.

Step 5 Fill the empty space.

  • In our example, we want to find the number to fill in the blank in 4_ × _ that makes the answer as large as possible but still less than or equal to 245. In this case, the answer is 5 . 45 × 5 = 225, while 46 × 6 = 276.

Step 6 Continue, using your

  • Continuing from our example, we would subtract 225 from 245 to get 20. Next, we would drop down the next pair of digits, 00, to make 2000. Doubling the numbers above the radical sign, we get 25 × 2 = 50. Solving for the blank in 50_ × _ =/< 2,000, we get 3 . At this point, we have "253" above the radical sign — repeating this process once again, we get a 9 as our next digit.

Step 7 Move the decimal point up from your original

  • In our example, the number under the radical sign is 6.45, so we would simply slide the point up and place it between the 2 and 5 digits of our answer, giving us 2.539 .

Quickly Estimating Imperfect Squares

Step 1 Find non-perfect squares by estimating.

  • For example, let's say we need to find the square root of 40. Since we've memorized our perfect squares, we can say that 40 is in between 6 2 and 7 2 , or 36 and 49. Since 40 is greater than 6 2 , its square root will be greater than 6, and since it is less than 7 2 , its square root will be less than 7. 40 is a little closer to 36 than it is to 49, so the answer will probably be a little closer to 6. In the next few steps, we'll narrow our answer down.

Step 2 Estimate the square root to one decimal point.

  • In our example problem, a reasonable estimate for the square root of 40 might be 6.4 , since we know from above that the answer is probably a little closer to 6 than it is to 7.

Step 3 Multiply your estimate by itself.

  • Multiply 6.4 by itself to get 6.4 × 6.4 = 40.96 , which is slightly higher than original number.
  • Next, since we over-shot our answer, we'll multiply the number one tenth less than our estimate above by itself and to get 6.3 × 6.3 = 39.69 . This is slightly lower than our original number. This means that the square root of 40 is somewhere between 6.3 and 6.4 . Additionally, since 39.69 is closer to 40 than 40.96, you know the square root will be closer to 6.3 than 6.4.

Step 4 Continue estimating as needed.

  • In our example, let's pick 6.33 for our two-decimal point estimate. Multiply 6.33 by itself to get 6.33 × 6.33 = 40.0689. Since this is slightly above our original number, we'll try a slightly lower number, like 6.32. 6.32 × 6.32 = 39.9424. This is slightly below our original number, so we know that the exact square root is between 6.33 and 6.32 . If we wanted to continue, we would keep using this same approach to get an answer that's continually more and more accurate.

Calculator, Practice Problems, and Answers

square root question with solution

Expert Q&A

David Jia

  • For quick solutions, use a calculator. Most modern calculators can instantly find square roots. Usually, all you need to do is to simply type in your number, then press the button with the square root symbol. To find the square root of 841, for example, you might press: 8, 4, 1, (√) and get an answer of 29 . [16] X Research source Thanks Helpful 0 Not Helpful 0

square root question with solution

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Calculate Frequency

  • ↑ David Jia. Academic Tutor. Expert Interview. 14 January 2021.
  • ↑ https://virtualnerd.com/algebra-foundations/powers-square-roots/powers-exponents/squaring-a-number
  • ↑ https://www.mathsisfun.com/square-root.html
  • ↑ http://virtualnerd.com/algebra-1/algebra-foundations/powers-square-roots/square-roots/square-root-estimation
  • ↑ https://www.cuemath.com/algebra/perfect-squares/
  • ↑ https://www.khanacademy.org/math/algebra/rational-exponents-and-radicals/alg1-simplify-square-roots/a/simplifying-square-roots-review
  • ↑ https://math.libretexts.org/Courses/Monroe_Community_College/MTH_165_College_Algebra_MTH_175_Precalculus/00%3A_Preliminary_Topics_for_College_Algebra/0.03%3A_Review_-_Radicals_(Square_Roots)
  • ↑ http://www.homeschoolmath.net/teaching/square-root-algorithm.php
  • ↑ https://www.cuemath.com/algebra/square-root-by-long-division-method/
  • ↑ https://virtualnerd.com/algebra-1/algebra-foundations/powers-square-roots/square-roots/square-root-estimation
  • ↑ http://www.math.com/students/calculators/source/square-root.htm

About This Article

David Jia

To solve square root problems, understand that you are finding the number that, when multiplied by itself, equals the number in the square root. For quick recall, memorize the first 10-12 perfect squares, so that you recognize the square root of numbers like 9, 25, 49, or 121. If possible, break the number under the square root into individual perfect squares. For example, √(900) can be broken into √(9) × √(100), and √(100) can be broken into √(25) × √(4), reducing the problem to √(9) × √(25) × √(4), or 3 x 5 x 2 for an answer of 30. If you want to learn how to estimate imperfect square roots, keep reading the article! Did this summary help you? Yes No

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  • 9.1 Solve Quadratic Equations Using the Square Root Property
  • Introduction
  • 1.1 Use the Language of Algebra
  • 1.2 Integers
  • 1.3 Fractions
  • 1.4 Decimals
  • 1.5 Properties of Real Numbers
  • Key Concepts
  • Review Exercises
  • Practice Test
  • 2.1 Use a General Strategy to Solve Linear Equations
  • 2.2 Use a Problem Solving Strategy
  • 2.3 Solve a Formula for a Specific Variable
  • 2.4 Solve Mixture and Uniform Motion Applications
  • 2.5 Solve Linear Inequalities
  • 2.6 Solve Compound Inequalities
  • 2.7 Solve Absolute Value Inequalities
  • 3.1 Graph Linear Equations in Two Variables
  • 3.2 Slope of a Line
  • 3.3 Find the Equation of a Line
  • 3.4 Graph Linear Inequalities in Two Variables
  • 3.5 Relations and Functions
  • 3.6 Graphs of Functions
  • 4.1 Solve Systems of Linear Equations with Two Variables
  • 4.2 Solve Applications with Systems of Equations
  • 4.3 Solve Mixture Applications with Systems of Equations
  • 4.4 Solve Systems of Equations with Three Variables
  • 4.5 Solve Systems of Equations Using Matrices
  • 4.6 Solve Systems of Equations Using Determinants
  • 4.7 Graphing Systems of Linear Inequalities
  • 5.1 Add and Subtract Polynomials
  • 5.2 Properties of Exponents and Scientific Notation
  • 5.3 Multiply Polynomials
  • 5.4 Dividing Polynomials
  • Introduction to Factoring
  • 6.1 Greatest Common Factor and Factor by Grouping
  • 6.2 Factor Trinomials
  • 6.3 Factor Special Products
  • 6.4 General Strategy for Factoring Polynomials
  • 6.5 Polynomial Equations
  • 7.1 Multiply and Divide Rational Expressions
  • 7.2 Add and Subtract Rational Expressions
  • 7.3 Simplify Complex Rational Expressions
  • 7.4 Solve Rational Equations
  • 7.5 Solve Applications with Rational Equations
  • 7.6 Solve Rational Inequalities
  • 8.1 Simplify Expressions with Roots
  • 8.2 Simplify Radical Expressions
  • 8.3 Simplify Rational Exponents
  • 8.4 Add, Subtract, and Multiply Radical Expressions
  • 8.5 Divide Radical Expressions
  • 8.6 Solve Radical Equations
  • 8.7 Use Radicals in Functions
  • 8.8 Use the Complex Number System
  • 9.2 Solve Quadratic Equations by Completing the Square
  • 9.3 Solve Quadratic Equations Using the Quadratic Formula
  • 9.4 Solve Equations in Quadratic Form
  • 9.5 Solve Applications of Quadratic Equations
  • 9.6 Graph Quadratic Functions Using Properties
  • 9.7 Graph Quadratic Functions Using Transformations
  • 9.8 Solve Quadratic Inequalities
  • 10.1 Finding Composite and Inverse Functions
  • 10.2 Evaluate and Graph Exponential Functions
  • 10.3 Evaluate and Graph Logarithmic Functions
  • 10.4 Use the Properties of Logarithms
  • 10.5 Solve Exponential and Logarithmic Equations
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  • 12.3 Geometric Sequences and Series
  • 12.4 Binomial Theorem

Learning Objectives

By the end of this section, you will be able to:

  • Solve quadratic equations of the form a x 2 = k a x 2 = k using the Square Root Property
  • Solve quadratic equations of the form a ( x – h ) 2 = k a ( x – h ) 2 = k using the Square Root Property

Be Prepared 9.1

Before you get started, take this readiness quiz.

Simplify: 128 . 128 . If you missed this problem, review Example 8.13 .

Be Prepared 9.2

Simplify: 32 5 32 5 . If you missed this problem, review Example 8.50 .

Be Prepared 9.3

Factor: 9 x 2 − 12 x + 4 9 x 2 − 12 x + 4 . If you missed this problem, review Example 6.23 .

A quadratic equation is an equation of the form ax 2 + bx + c = 0, where a ≠ 0 a ≠ 0 . Quadratic equations differ from linear equations by including a quadratic term with the variable raised to the second power of the form ax 2 . We use different methods to solve quadratic equations than linear equations, because just adding, subtracting, multiplying, and dividing terms will not isolate the variable.

We have seen that some quadratic equations can be solved by factoring. In this chapter, we will learn three other methods to use in case a quadratic equation cannot be factored.

Solve Quadratic Equations of the form a x 2 = k a x 2 = k using the Square Root Property

We have already solved some quadratic equations by factoring. Let’s review how we used factoring to solve the quadratic equation x 2 = 9.

We can easily use factoring to find the solutions of similar equations, like x 2 = 16 and x 2 = 25, because 16 and 25 are perfect squares. In each case, we would get two solutions, x = 4 , x = −4 x = 4 , x = −4 and x = 5 , x = −5 . x = 5 , x = −5 .

But what happens when we have an equation like x 2 = 7? Since 7 is not a perfect square, we cannot solve the equation by factoring.

Previously we learned that since 169 is the square of 13, we can also say that 13 is a square root of 169. Also, (−13) 2 = 169, so −13 is also a square root of 169. Therefore, both 13 and −13 are square roots of 169. So, every positive number has two square roots—one positive and one negative. We earlier defined the square root of a number in this way:

Since these equations are all of the form x 2 = k , the square root definition tells us the solutions are the two square roots of k . This leads to the Square Root Property .

Square Root Property

If x 2 = k , then

Notice that the Square Root Property gives two solutions to an equation of the form x 2 = k , the principal square root of k k and its opposite. We could also write the solution as x = ± k . x = ± k . We read this as x equals positive or negative the square root of k .

Now we will solve the equation x 2 = 9 again, this time using the Square Root Property.

What happens when the constant is not a perfect square? Let’s use the Square Root Property to solve the equation x 2 = 7.

We cannot simplify 7 7 , so we leave the answer as a radical.

Example 9.1

How to solve a quadratic equation of the form ax 2 = k using the square root property.

Solve: x 2 − 50 = 0 . x 2 − 50 = 0 .

Solve: x 2 − 48 = 0 . x 2 − 48 = 0 .

Solve: y 2 − 27 = 0 . y 2 − 27 = 0 .

The steps to take to use the Square Root Property to solve a quadratic equation are listed here.

Solve a quadratic equation using the square root property.

  • Step 1. Isolate the quadratic term and make its coefficient one.
  • Step 2. Use Square Root Property.
  • Step 3. Simplify the radical.
  • Step 4. Check the solutions.

In order to use the Square Root Property, the coefficient of the variable term must equal one. In the next example, we must divide both sides of the equation by the coefficient 3 before using the Square Root Property.

Example 9.2

Solve: 3 z 2 = 108 . 3 z 2 = 108 .

Solve: 2 x 2 = 98 . 2 x 2 = 98 .

Solve: 5 m 2 = 80 . 5 m 2 = 80 .

The Square Root Property states ‘If x 2 = k x 2 = k ,’ What will happen if k < 0 ? k < 0 ? This will be the case in the next example.

Example 9.3

Solve: x 2 + 72 = 0 x 2 + 72 = 0 .

Solve: c 2 + 12 = 0 . c 2 + 12 = 0 .

Solve: q 2 + 24 = 0 . q 2 + 24 = 0 .

Our method also works when fractions occur in the equation; we solve as any equation with fractions. In the next example, we first isolate the quadratic term, and then make the coefficient equal to one.

Example 9.4

Solve: 2 3 u 2 + 5 = 17 . 2 3 u 2 + 5 = 17 .

Solve: 1 2 x 2 + 4 = 24 . 1 2 x 2 + 4 = 24 .

Solve: 3 4 y 2 − 3 = 18 . 3 4 y 2 − 3 = 18 .

The solutions to some equations may have fractions inside the radicals. When this happens, we must rationalize the denominator .

Example 9.5

Solve: 2 x 2 − 8 = 41 . 2 x 2 − 8 = 41 .

Solve: 5 r 2 − 2 = 34 . 5 r 2 − 2 = 34 .

Try It 9.10

Solve: 3 t 2 + 6 = 70 . 3 t 2 + 6 = 70 .

Solve Quadratic Equations of the Form a ( x − h ) 2 = k Using the Square Root Property

We can use the Square Root Property to solve an equation of the form a ( x − h ) 2 = k as well. Notice that the quadratic term, x , in the original form ax 2 = k is replaced with ( x − h ).

The first step, like before, is to isolate the term that has the variable squared. In this case, a binomial is being squared. Once the binomial is isolated, by dividing each side by the coefficient of a , then the Square Root Property can be used on ( x − h ) 2 .

Example 9.6

Solve: 4 ( y − 7 ) 2 = 48 . 4 ( y − 7 ) 2 = 48 .

Try It 9.11

Solve: 3 ( a − 3 ) 2 = 54 . 3 ( a − 3 ) 2 = 54 .

Try It 9.12

Solve: 2 ( b + 2 ) 2 = 80 . 2 ( b + 2 ) 2 = 80 .

Remember when we take the square root of a fraction, we can take the square root of the numerator and denominator separately.

Example 9.7

Solve: ( x − 1 3 ) 2 = 5 9 . ( x − 1 3 ) 2 = 5 9 .

Try It 9.13

Solve: ( x − 1 2 ) 2 = 5 4 . ( x − 1 2 ) 2 = 5 4 .

Try It 9.14

Solve: ( y + 3 4 ) 2 = 7 16 . ( y + 3 4 ) 2 = 7 16 .

We will start the solution to the next example by isolating the binomial term.

Example 9.8

Solve: 2 ( x − 2 ) 2 + 3 = 57 . 2 ( x − 2 ) 2 + 3 = 57 .

Try It 9.15

Solve: 5 ( a − 5 ) 2 + 4 = 104 . 5 ( a − 5 ) 2 + 4 = 104 .

Try It 9.16

Solve: 3 ( b + 3 ) 2 − 8 = 88 . 3 ( b + 3 ) 2 − 8 = 88 .

Sometimes the solutions are complex numbers.

Example 9.9

Solve: ( 2 x − 3 ) 2 = −12 . ( 2 x − 3 ) 2 = −12 .

Try It 9.17

Solve: ( 3 r + 4 ) 2 = −8 . ( 3 r + 4 ) 2 = −8 .

Try It 9.18

Solve: ( 2 t − 8 ) 2 = −10 . ( 2 t − 8 ) 2 = −10 .

The left sides of the equations in the next two examples do not seem to be of the form a ( x − h ) 2 . But they are perfect square trinomials, so we will factor to put them in the form we need.

Example 9.10

Solve: 4 n 2 + 4 n + 1 = 16 . 4 n 2 + 4 n + 1 = 16 .

We notice the left side of the equation is a perfect square trinomial. We will factor it first.

Try It 9.19

Solve: 9 m 2 − 12 m + 4 = 25 . 9 m 2 − 12 m + 4 = 25 .

Try It 9.20

Solve: 16 n 2 + 40 n + 25 = 4 . 16 n 2 + 40 n + 25 = 4 .

Access this online resource for additional instruction and practice with using the Square Root Property to solve quadratic equations.

  • Solving Quadratic Equations: The Square Root Property
  • Using the Square Root Property to Solve Quadratic Equations

Section 9.1 Exercises

Practice makes perfect.

Solve Quadratic Equations of the Form ax 2 = k Using the Square Root Property

In the following exercises, solve each equation.

a 2 = 49 a 2 = 49

b 2 = 144 b 2 = 144

r 2 − 24 = 0 r 2 − 24 = 0

t 2 − 75 = 0 t 2 − 75 = 0

u 2 − 300 = 0 u 2 − 300 = 0

v 2 − 80 = 0 v 2 − 80 = 0

4 m 2 = 36 4 m 2 = 36

3 n 2 = 48 3 n 2 = 48

4 3 x 2 = 48 4 3 x 2 = 48

5 3 y 2 = 60 5 3 y 2 = 60

x 2 + 25 = 0 x 2 + 25 = 0

y 2 + 64 = 0 y 2 + 64 = 0

x 2 + 63 = 0 x 2 + 63 = 0

y 2 + 45 = 0 y 2 + 45 = 0

4 3 x 2 + 2 = 110 4 3 x 2 + 2 = 110

2 3 y 2 − 8 = −2 2 3 y 2 − 8 = −2

2 5 a 2 + 3 = 11 2 5 a 2 + 3 = 11

3 2 b 2 − 7 = 41 3 2 b 2 − 7 = 41

7 p 2 + 10 = 26 7 p 2 + 10 = 26

2 q 2 + 5 = 30 2 q 2 + 5 = 30

5 y 2 − 7 = 25 5 y 2 − 7 = 25

3 x 2 − 8 = 46 3 x 2 − 8 = 46

( u − 6 ) 2 = 64 ( u − 6 ) 2 = 64

( v + 10 ) 2 = 121 ( v + 10 ) 2 = 121

( m − 6 ) 2 = 20 ( m − 6 ) 2 = 20

( n + 5 ) 2 = 32 ( n + 5 ) 2 = 32

( r − 1 2 ) 2 = 3 4 ( r − 1 2 ) 2 = 3 4

( x + 1 5 ) 2 = 7 25 ( x + 1 5 ) 2 = 7 25

( y + 2 3 ) 2 = 8 81 ( y + 2 3 ) 2 = 8 81

( t − 5 6 ) 2 = 11 25 ( t − 5 6 ) 2 = 11 25

( a − 7 ) 2 + 5 = 55 ( a − 7 ) 2 + 5 = 55

( b − 1 ) 2 − 9 = 39 ( b − 1 ) 2 − 9 = 39

4 ( x + 3 ) 2 − 5 = 27 4 ( x + 3 ) 2 − 5 = 27

5 ( x + 3 ) 2 − 7 = 68 5 ( x + 3 ) 2 − 7 = 68

( 5 c + 1 ) 2 = −27 ( 5 c + 1 ) 2 = −27

( 8 d − 6 ) 2 = −24 ( 8 d − 6 ) 2 = −24

( 4 x − 3 ) 2 + 11 = −17 ( 4 x − 3 ) 2 + 11 = −17

( 2 y + 1 ) 2 − 5 = −23 ( 2 y + 1 ) 2 − 5 = −23

m 2 − 4 m + 4 = 8 m 2 − 4 m + 4 = 8

n 2 + 8 n + 16 = 27 n 2 + 8 n + 16 = 27

x 2 − 6 x + 9 = 12 x 2 − 6 x + 9 = 12

y 2 + 12 y + 36 = 32 y 2 + 12 y + 36 = 32

25 x 2 − 30 x + 9 = 36 25 x 2 − 30 x + 9 = 36

9 y 2 + 12 y + 4 = 9 9 y 2 + 12 y + 4 = 9

36 x 2 − 24 x + 4 = 81 36 x 2 − 24 x + 4 = 81

64 x 2 + 144 x + 81 = 25 64 x 2 + 144 x + 81 = 25

Mixed Practice

In the following exercises, solve using the Square Root Property.

2 r 2 = 32 2 r 2 = 32

4 t 2 = 16 4 t 2 = 16

( a − 4 ) 2 = 28 ( a − 4 ) 2 = 28

( b + 7 ) 2 = 8 ( b + 7 ) 2 = 8

9 w 2 − 24 w + 16 = 1 9 w 2 − 24 w + 16 = 1

4 z 2 + 4 z + 1 = 49 4 z 2 + 4 z + 1 = 49

a 2 − 18 = 0 a 2 − 18 = 0

b 2 − 108 = 0 b 2 − 108 = 0

( p − 1 3 ) 2 = 7 9 ( p − 1 3 ) 2 = 7 9

( q − 3 5 ) 2 = 3 4 ( q − 3 5 ) 2 = 3 4

m 2 + 12 = 0 m 2 + 12 = 0

n 2 + 48 = 0 . n 2 + 48 = 0 .

u 2 − 14 u + 49 = 72 u 2 − 14 u + 49 = 72

v 2 + 18 v + 81 = 50 v 2 + 18 v + 81 = 50

( m − 4 ) 2 + 3 = 15 ( m − 4 ) 2 + 3 = 15

( n − 7 ) 2 − 8 = 64 ( n − 7 ) 2 − 8 = 64

( x + 5 ) 2 = 4 ( x + 5 ) 2 = 4

( y − 4 ) 2 = 64 ( y − 4 ) 2 = 64

6 c 2 + 4 = 29 6 c 2 + 4 = 29

2 d 2 − 4 = 77 2 d 2 − 4 = 77

( x − 6 ) 2 + 7 = 3 ( x − 6 ) 2 + 7 = 3

( y − 4 ) 2 + 10 = 9 ( y − 4 ) 2 + 10 = 9

Writing Exercises

In your own words, explain the Square Root Property.

In your own words, explain how to use the Square Root Property to solve the quadratic equation ( x + 2 ) 2 = 16 ( x + 2 ) 2 = 16 .

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

Choose how would you respond to the statement “I can solve quadratic equations of the form a times the square of x minus h equals k using the Square Root Property.” “Confidently,” “with some help,” or “No, I don’t get it.”

ⓑ If most of your checks were:

…confidently. Congratulations! You have achieved the objectives in this section. Reflect on the study skills you used so that you can continue to use them. What did you do to become confident of your ability to do these things? Be specific.

…with some help. This must be addressed quickly because topics you do not master become potholes in your road to success. In math every topic builds upon previous work. It is important to make sure you have a strong foundation before you move on. Whom can you ask for help?Your fellow classmates and instructor are good resources. Is there a place on campus where math tutors are available? Can your study skills be improved?

…no - I don’t get it! This is a warning sign and you must not ignore it. You should get help right away or you will quickly be overwhelmed. See your instructor as soon as you can to discuss your situation. Together you can come up with a plan to get you the help you need.

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Want to cite, share, or modify this book? This book uses the Creative Commons Attribution License and you must attribute OpenStax.

Access for free at https://openstax.org/books/intermediate-algebra-2e/pages/1-introduction
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  • Publisher/website: OpenStax
  • Book title: Intermediate Algebra 2e
  • Publication date: May 6, 2020
  • Location: Houston, Texas
  • Book URL: https://openstax.org/books/intermediate-algebra-2e/pages/1-introduction
  • Section URL: https://openstax.org/books/intermediate-algebra-2e/pages/9-1-solve-quadratic-equations-using-the-square-root-property

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  • Square Root Questions

ffImage

What is Square Root?

A square root is a value that, when multiplied by itself, gives the original number.

How to Find the Square Root?

We can use two methods to find square roots - Prime factorization and the Long division method.

What are Squares and Square Roots?

Just the opposite method of squaring a number is the square root. If a number n, such as n 2 , is squared, then the square root of n 2 is equal to the original number n.

How to Square a Number?

We need to multiply the number by itself to find the square of a number.

For example, 3 multiplied by 3 is equal to 9.

Square of 4 is: 4 multiplied by 4 = 4 × 4 = 16.

Square Root Problems and Answers

Q1: Can you find out the number of numbers lying between the squares of these following pairs of numbers?

(i) 25 and 26

(ii) 99 and 100

Solution - As we know, between n 2 and (n + 1) 2 , the number of non–perfect square numbers are 2n.

(i) Between 252 and 262 there are 2 × 25 = 50 natural numbers.

(ii) Between 992 and 1002 there are 2 × 99 = 198 natural numbers.

Q2: Write a Pythagorean triplet whose one of the required member is:

We know, for any natural number m, 2m, m 2 – 1, m 2 + 1 is a Pythagorean triplet.

⇒ m = 6/2 = 3

m 2 – 1 = 3 2 – 1 = 9 – 1 = 8

m 2 + 1 = 3 2 + 1 = 9 + 1 = 10

Therefore, (6, 8, 10) is a Pythagorean triplet.

(ii) 2m = 14

⇒ m = 14/2 = 7

m 2 – 1 = 72 – 1 = 49 – 1 = 48

m 2 + 1 = 72 + 1 = 49 + 1 = 50

(14, 48, 50) is not a Pythagorean triplet.

(iii) 2m = 16

⇒ m = 16/2 = 8

m 2 – 1 = 8 2 – 1 = 64 – 1 = 63

m 2 + 1 = 8 2 + 1 = 64 + 1 = 65

Therefore, (16, 63, 65) is a Pythagorean triplet.

(iv) 2m = 18

⇒ m = 18/2 = 9

m 2 – 1 = 92 – 1 = 81 – 1 = 80

m 2 + 1 = 92 + 1 = 81 + 1 = 82

Therefore, (18, 80, 82) is a Pythagorean triplet.

Q3: (n + 1) 2 - n 2 = ?

(n + 1) 2 - n 2

= (n 2 + 2n + 1) – n 2

Q4: State that the number 121 is the sum of 11 odd natural numbers.

Solution - As 121 = 112

We recognize that n2 is the sum of the first n odd natural numbers.

It shows that 121 = the sum of the first 11 odd natural numbers.

= 1 + 3 + 5+ 7 + 9 + 11 +13 + 15 + 17 + 19 + 21

Q5: Use the identity and find the square of 189.

(a – b) 2 = a 2 – 2ab + b 2

Solution - 189 = (200 – 11) 2

= 40000 – 2 × 200 × 11 + 112

= 40000 – 4400 + 121

Q6: Find out the square root of 625 using the mathematical identity as stated:  (a + b) 2 = a 2 + b 2 + 2ab?

Solution - (625) 2

= (600 + 25) 2

= 6002 + 2 × 600 × 25 + 252

= 360000 + 30000 + 625

Properties of Square Roots

A square root function is defined in mathematics as a one-to-one function that takes as an input a positive number and returns the square root of the given input number.

The following are some of the essential properties of the square root:

If a number is a perfect square number, a perfect square root exists.

It may have a square root if a number ends with an even number of zeros (0's).

It is possible to multiply the two square root values. For instance, √3 can be multiplied by √2, then √6 should be the result.

If two same square roots are multiplied, then a radical number should be the product. This implies that a non-square root number is a product. When √7 is multiplied by √7, for example, the result obtained is 7.

It does not define the square root of any negative numbers. And no negative can be the perfect square.

If a number ends with 2, 3, 7, or 8 (in the digit of the unit), the perfect square root will not exist.

If in the unit digit, a number ends with 1, 4, 5, 6, or 9, the number would have a square root.

Using the prime factorization process, the square root of a perfect square number is simple to measure. For example -

Square Root By Prime Factorisation

With the help of an example, let us understand this concept:

Example 1: Solve √10 to 2 decimal places.

Step 1: Choose any two perfect square roots between which you feel your number may fall.

We know that 22 = 4; 32 = 9, 42 = 16 and 52 = 25

Now, choose 3 and 4 (as √10 lies between these 2 numbers)

Step 2: Divide the given number into one of the square roots chosen.

Divide 10 by 3.

=> 10/3 = 3.33 (round off answer at 2 places)

Step 3: Find the root average and the product of the step above, i.e.

(3 + 3.33)/2 = 3.1667

Verify: 3.1667 × 3.1667 = 10.0279 (Not required)

Repeat step 2 and step 3

Now 10/3.1667 = 3.1579

Average of 3.1667 and 3.1579.

(3.1667+3.1579)/2 = 3.1623

Verify: 3.1623 × 3.1623 = 10.0001 (more accurate)

Stop the process.

Example 2: Find the square roots of whole numbers from 1 to 100 that are perfect squares.

Solution - The perfect squares are - 1, 4, 9, 16, 25, 36, 49, 64, 81, 100.

Example 3: What is:

The square root of 2.

The square root of 3.

The square root of 4.

The square root of 5

Solution - Use a square root list, we have

Value of root 2 i.e. √2 = 1.4142.

Value of root 3 i.e. √3 = 1.7321.

Value of root 4 i.e. √4 = 2.

Value of root 5 i.e. √5 = 2.2361.

arrow-right

FAQs on Square Root Questions

Question 1. Define Square Numbers and Square Roots.

Answer: Square numbers can be defined as the numbers that are produced when a number is multiplied by itself. For example, if n is a number and it is multiplied by itself, then the square of n can be given as n 2 . Another example is the square of 10, which is 10 2 = 10 x 10 = 100.

Further, the square root of a number can be explained as the value which when multiplied by itself gives the original number. This value can be represented by the symbol ‘√.’ For example, the square root of 25 is √25 = 5.

Question 2. What do you Understand by Perfect Squares? Provide Examples with Your Answer.

Answer: Perfect squares can be defined as numbers in which the square root of the number provides a whole number. For example, 9 is a perfect square number because its root is a whole number, which means that √9 = 3.

Question 3. What is an Imperfect Square? Provide Examples with Your Answer.

Answer: An imperfect square can be defined as a number in which the square root of the number is a fraction. The value is generated by taking the square root of the imperfect square can also be non-terminating. For example, 3 is an imperfect square because the root of this number is 1.73205080757. This number is a fraction.

Question 4. What is the Main Difference Between a Square and Square Roots?

Answer: The square root of a number provides the root of a particular number that was squared. This is the main difference between a square number and square roots.

Question 5. How to Solve the Square Root Equation.

Answer: We need to follow the below steps to solve the square root equation: Isolate the square to one of the sides (L.H.S or R.H.S). Now solve the remaining equation. Square both sides of the given equation. With examples, let us understand the steps.

Question 6. Is the Square Root of a Negative Number a Whole Number?

Answer: No, negative numbers shouldn't have a square root, as per the square root concept. If we multiply two negative numbers, a positive number will always be the product. Negative number square roots expressed as multiples of I (imaginary numbers).

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4.8: The Square Root Function

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In this section we turn our attention to the square root function, the function defined by the equation

\[\begin{array}{c} {f(x)= \sqrt{x}}\\ \end{array} \nonumber \]

We begin the section by drawing the graph of the function, then we address the domain and range. After that, we’ll investigate a number of different transformations of the function.

The Graph of the Square Root Function

Let’s create a table of points that satisfy the equation of the function, then plot the points from the table on a Cartesian coordinate system on graph paper. We’ll continue creating and plotting points until we are convinced of the eventual shape of the graph.

We know we cannot take the square root of a negative number. Therefore, we don’t want to put any negative x -values in our table. To further simplify our computations, let’s use numbers whose square root is easily calculated. This brings to mind perfect squares such as 0, 1, 4, 9, and so on. We’ve placed these numbers as x -values in the table in Figure 1 (b), then calculated the square root of each. In Figure 1 (a), you see each of the points from the table plotted as a solid dot. If we continue to add points to the table, plot them, the graph will eventually fill in and take the shape of the solid curve shown in Figure 1 (c).

Screen Shot 2019-05-24 at 3.15.38 PM.png

The point plotting approach used to draw the graph of \(f(x) = \sqrt{x}\) in Figure 1 is a tested and familiar procedure. However, a more sophisticated approach involves the theory of inverses developed in the previous chapter.

In a sense, taking the square root is the “inverse” of squaring. Well, not quite, as the squaring function \(f(x) = x^2\) in Figure 2 (a) fails the horizontal line test and is not one-to-one. However, if we limit the domain of the squaring function, then the graph of \(f(x) = x^2\) in Figure 2 (b), where \(x \ge 0\), does pass the horizontal line test and is one-to-one. Therefore, the graph of \(f(x) = x^2\), \(x \ge 0\), has an inverse, and the graph of its inverse is found by reflecting the graph of \(f(x) = x^2\), \(x \ge 0\), across the line y = x (see Figure 2 (c)).

Screen Shot 2019-05-24 at 3.22.52 PM.png

To find the equation of the inverse, recall that the procedure requires that we switch the roles of x and y , then solve the resulting equation for y . Thus, first write \(f(x) = x^2\), \(x \ge 0\), in the form

\[\begin{array}{c} {y = x^2, x \ge 0}\\ \nonumber \end{array} \nonumber \]

Next, switch x and y .

\[\begin{array}{c} {x = y^2, y \ge 0}\\ \end{array} \nonumber \]

When we solve this last equation for y , we get two solutions,

\[\begin{array}{c} {y = \pm\sqrt{x}}\\ \end{array} \nonumber \]

However, in equation (2) , note that y must be greater than or equal to zero. Hence, we must choose the nonnegative answer in equation (3) , so the inverse of \(f(x) = x^2\), \(x \ge 0\), has equation

\[\begin{array}{c} {f^{−1}(x) = \sqrt{x}}\\ \nonumber \end{array} \nonumber \]

This is the equation of the reflection of the graph of \(f(x) = x^2\), \(x \ge 0\), that is pictured in Figure 2 (c). Note the exact agreement with the graph of the square root function in Figure 1 (c).

The sequence of graphs in Figure 2 also help us identify the domain and range of the square root function.

  • In Figure 2 (a), the parabola opens outward indefinitely, both left and right. Consequently, the domain is \(D_{f} = (−\infty, \infty)\), or all real numbers. Also, the graph has vertex at the origin and opens upward indefinitely, so the range is \(R_{f} = [0, \infty)\).
  • In Figure 2 (b), we restricted the domain. Thus, the graph of \(f(x) = x^2\), \(x \ge 0\), now has domain \(D_{f} = [0, \infty)\). The range is unchanged and is \(R_{f} = [0, \infty)\).
  • In Figure 2 (c), we’ve reflected the graph of \(f(x) = x^2\), \(x \ge 0\), across the line y = x to obtain the graph of \(f^{−1}(x) = \sqrt{x}\). Because we’ve interchanged the role of x and y, the domain of the square root function must equal the range of \(f(x) = x^2\), \(x \ge 0\). That is, \(D_{f^{−1}} =[0,\infty)\).Similarly,the range of the square root function must equal the domain of \(f(x) = x^2\), \(x \ge 0\). Hence, \(R_{f^{−1}} = [0,\infty)\).

Of course, we can also determine the domain and range of the square root function by projecting all points on the graph onto the x - and y -axes, as shown in Figures 3 (a) and (b), respectively.

Screen Shot 2019-05-24 at 3.49.15 PM.png

Some might object to the range, asking “How do we know that the graph of the square root function picture in Figure 3 (b) rises indefinitely?” Again, the answer lies in the sequence of graphs in Figure 2 . In Figure 2 (c), note that the graph of \(f(x) = x^2\), \(x \ge 0\), opens indefinitely to the right as the graph rises to infinity. Hence, after reflecting this graph across the line y = x , the resulting graph must rise upward indefinitely as it moves to the right. Thus, the range of the square root function is \([0, \infty)\).

Translations

If we shift the graph of \(y = \sqrt{x}\) right and left, or up and down, the domain and/or range are affected.

Example \(\PageIndex{4}\)

Sketch the graph of \(f(x) = \sqrt{x−2}\). Use your graph to determine the domain and range.

We know that the basic equation \(y=\sqrt{x}\) has the graph shown in Figures 1 (c). If we replace x with x − 2, the basic equation \(y=\sqrt{x}\) becomes \(f(x) = \sqrt{x−2}\) . From our previous work with geometric transformations, we know that this will shift the graph two units to the right, as shown in Figures 4 (a) and (b).

Screen Shot 2019-05-24 at 4.00.39 PM.png

To find the domain, we project each point on the graph of f onto the x-axis, as shown in Figure 4 (a). Note that all points to the right of or including 2 are shaded on the x-axis. Consequently, the domain of f is

Domain = \([2, \infty)\) = {x: \(x \ge 0\)}

As there has been no shift in the vertical direction, the range remains the same. To find the range, we project each point on the graph onto the y-axis, as shown in Figure 4 (b). Note that all points at and above zero are shaded on the y-axis. Thus, the range of f is

Range = \([0,\infty)\)= {y: \(y \ge 0\)}.

We can find the domain of this function algebraically by examining its defining equation \(f(x) = \sqrt{x−2}\). We understand that we cannot take the square root of a negative number. Therefore, the expression under the radical must be nonnegative (positive or zero). That is,

\(x − 2 \ge 0\).

Solving this inequality for x ,

\(x \ge 2\).

Thus, the domain of f is Domain = \([2, \infty)\), which matches the graphical solution above.

Let’s look at another example.

Example \(\PageIndex{5}\)

Sketch the graph of \(f (x) = \sqrt{x + 4} + 2\). Use your graph to determine the domain and range of f.

Again, we know that the basic equation \(y=\sqrt{x}\) has the graph shown in Figure 1 (c). If we replace x with x +4, the basic equation \(y=\sqrt{x}\) becomes \(y=\sqrt{x+4}\) . From our previous work with geometric transformations, we know that this will shift the graph of \(y=\sqrt{x}\) four units to the left, as shown in Figure 5 (a).

If we know add 2 to the equation \(y=\sqrt{x+4}\) to produce the equation \(y=\sqrt{x+4} + 2\), this will shift the graph of \(y=\sqrt{x+4}\) two units upward, as shown in Figure 5 (b).

Screen Shot 2019-06-14 at 3.41.30 PM.png

To identify th domain of the \(f (x) = \sqrt{x + 4} + 2\), we project all points on the graph of f onto the x-axis, as shown in Figure 6 (a). Note that all points to the right of or including − 4 are shaded on the x -axis. Thus, the domain of \(f (x) = \sqrt{x + 4} + 2\) is

Domain = \([−4, \infty)\) = {x: \(x \ge −4\)}

square root question with solution

Similarly, to find the range of f , project all points on the graph of f onto the y -axis, as shown in Figure 6 (b). Note that all points on the y -axis greater than or including 2 are shaded. Consequently, the range of f is

Range = \([2, \infty)\) = {y: \(y \ge 2\)}

We can also find the domain of f algebraically by examining the equation \(f (x) = \sqrt{x + 4} + 2\). We cannot take the square root of a negative number, so the expression under the radical must be nonnegative (zero or positive). Consequently,

\(x + 4 \ge 0\).

\(x \ge −4\).

Thus, the domain of f is Domain = \([−4,\infty)\), which matches the graphical solution presented above.

Reflections

If we start with the basic equation \(y = \sqrt{x}\), then replace x with −x, then the graph of the resulting equation \(y = \sqrt{−x}\) is captured by reflecting the graph of \(y = \sqrt{x}\) (see Figure 1 (c)) horizontally across the y-axis. The graph of \(y = \sqrt{−x}\) is shown in Figure 7 (a).

Similarly, the graph of \(y = −\sqrt{x}\) would be a vertical reflection of the graph of \(y = \sqrt{x}\) across the x-axis, as shown in Figure 7 (b).

Screen Shot 2019-06-14 at 3.43.13 PM.png

More often than not, you will be asked to perform a reflection and a translation.

Example \(\PageIndex{6}\)

Sketch the graph of \(f(x) = \sqrt{4− x}\). Use the resulting graph to determine the domain and range of f.

First, rewrite the equation \(f(x) = \sqrt{4− x}\) as follows:

\(f(x) = \sqrt{−(x−4)}\)

Reflections First . It is usually more intuitive to perform reflections before translations.

With this thought in mind, we first sketch the graph of \(f(x) = \sqrt{−x}\) , which is a reflection of the graph of \(f(x) = \sqrt{x}\) across the y -axis. This is shown in Figure 8 (a).

Now, in \(f(x) = \sqrt{−x}\) replace x with x − 4 to obtain \(f(x) = \sqrt{−(x−4)}\). This shifts the graph of \(f(x) = \sqrt{−x}\) f our units to the right, as pictured in Figure 8 (b).

Screen Shot 2019-06-14 at 3.45.43 PM.png

To find the domain of the function \(f(x) = \sqrt{−(x−4)}\), or equivalently, \(f(x) = \sqrt{4−x}\) , project each point on the graph of f onto the x -axis, as shown in Figure 9 (a). Note that all real numbers less than or equal to 4 are shaded on the x -axis. Hence, the domain of f is

Domain = \((−\infty, 4]\) = {x: \(x \le 4\)}.

Similarly, to obtain the range of f, project each point on the graph of f onto they-axis, as shown in Figure 9 (b). Note that all real numbers greater than or equal to zero are shaded on the y-axis. Hence, the range of f is

Range = \([0,\infty)\) = {x: \(x \ge 0\)}.

We can also find the domain of the function f by examining the equation \(f(x) = \sqrt{4−x}\). We cannot take the square root of a negative number, so the expression under the radical must be nonnegative (zero or positive). Consequently,

\(4 − x \ge 0\).

Screen Shot 2019-06-14 at 3.46.52 PM.png

Solve this last inequality for x . First subtract 4 from both sides of the inequality, then multiply both sides of the resulting inequality by − 1. Of course, multiplying by a negative number reverses the inequality symbol.

\(−x \ge −4\)

\(x \le 4\)

Thus, the domain of f is {x: \(x \le 4\)}. In interval notation, Domain = \((−\infty, 4]\). This agree nicely with the graphical result found above.

More often than not, it will take a combination of your graphing calculator and a little algebraic manipulation to determine the domain of a square root function.

Example \(\PageIndex{7}\)

Sketch the graph of \(f(x) = \sqrt{5−2x}\) Use the graph and an algebraic technique to determine the domain of the function.

Load the function into Y1 in the Y= menu of your calculator, as shown in Figure 10 (a). Select 6: ZStandard from the ZOOM menu to produce the graph shown in Figure 10 (b).

Screen Shot 2019-06-14 at 3.48.53 PM.png

Look carefully at the graph in Figure 10 (b) and note that it’s difficult to tell if the graph comes all the way down to “touch” the x-axis near \(x \approx 2.5\). However, our previous experience with the square root function makes us believe that this is just an artifact of insufficient resolution on the calculator that is preventing the graph from “touching” the x-axis at \(x \approx 2.5\).

An algebraic approach will settle the issue. We can determine the domain of f by examine the equation \(f(x) = \sqrt{5 − 2x}\). Consequently, We cannot take the square root of a negative number, so the expression under the radical must be nonnegative (zero or positive).

\(5 − 2x \ge 0\).

Solve this last inequality for x . First, subtract 5 from both sides of the inequality.

\(−2x \ge −5\).

Next, divide both sides of this last inequality by −2. Remember that we must reverse the inequality the moment we divide by a negative number.

\(\frac{−2x}{−2} \le \frac{−5}{−2}\).

\(x \le \frac{5}{2}\).

Thus, the domain of f is {x: \(x \le \frac{5}{2}\)}. In interval notation, Domain = \((−\infty, \frac{5}{2}]\). This agree nicely with the graphical result found above.

Further introspection reveals that this argument also settles the issue of whether or not the graph “touches” the x -axis at \(x= \frac{5}{2}\). If you remain unconvinced, then substitute \(x=\frac{5}{2}\) in \(f(x) = \sqrt{5−2x}\) to see

\(f(\frac{5}{2})= \sqrt{5−2(\frac{5}{2})} =\sqrt{0} = 0\).

Thus, the graph of f “touches” the x-axis at the point \((\frac{5}{2}, 0)\).

In Exercise 1-10 , complete each of the following tasks:

  • Set up a coordinate system on a sheet of graph paper. Label and scale each axis.
  • Complete the table of points for the given function. Plot each of the points on your coordinate system, then use them to help draw the graph of the given function.
  • Use different colored pencils to project all points onto the x - and y -axes to determine the domain and range. Use interval notation to describe the do- main of the given function.

Exercise \(\PageIndex{1}\)

\(f(x) = −\sqrt{x}\)

Plot the points in the table and use them to help draw the graph.

Screen Shot 2019-05-28 at 11.25.20 PM.png

Project all points on the graph onto the x-axis to determine the domain: Domain = \([0, \infty)\). Project all points on the graph onto the y-axis to determine the range: Range = \((−\infty, 0]\).

Exercise \(\PageIndex{2}\)

\(f(x) = \sqrt{−x}\)

Exercise \(\PageIndex{3}\)

\(f(x)= \sqrt{x+2}\)

Screen Shot 2019-05-28 at 11.29.25 PM.png

Project all points on the graph onto the x-axis to determine the domain: Domain = \([ − 2, \infty)\). Project all points on the graph onto the y-axis to determine the range: Range = \([0, \infty)\).

Exercise \(\PageIndex{4}\)

\(f(x)= \sqrt{5−x}\)

Exercise \(\PageIndex{5}\)

\(f(x)= \sqrt{x}+2\)

Plot the points in the table and use them to draw the graph of f .

Screen Shot 2019-05-28 at 11.33.13 PM.png

Project all points on the graph onto the x-axis to determine the domain: Domain = \([0, \infty)\). Project all points on the graph onto the y-axis to determine the range: Range = \([2, \infty)\).

Exercise \(\PageIndex{6}\)

\(f(x)=\sqrt{x}−1\)

Exercise \(\PageIndex{7}\)

\(f(x)= \sqrt{x+3}+2\)

Screen Shot 2019-05-28 at 11.38.27 PM.png

Exercise \(\PageIndex{8}\)

\(f(x)= \sqrt{x−1}+3\)

Exercise \(\PageIndex{9}\)

\(f(x)= \sqrt{3−x}\)

Screen Shot 2019-05-28 at 11.42.01 PM.png

Project all points on the graph onto the x-axis to determine the domain: Domain = \(( − \infty, 3]\). Project all points on the graph onto the y-axis to determine the range: Range = \([0, \infty)\).

Exercise \(\PageIndex{10}\)

\(f(x)=−\sqrt{x+3}\)

In Exercises 11 - 20 , perform each of the following tasks.

  • Set up a coordinate system on a sheet of graph paper. Label and scale each axis. Remember to draw all lines with a ruler.
  • Use geometric transformations to draw the graph of the given function on your coordinate system without the use of a graphing calculator. Note: You may check your solution with your calculator, but you should be able to produce the graph without the use of your calculator.
  • Use different colored pencils to project the points on the graph of the function onto the x - and y -axes. Use interval notation to describe the domain and range of the function.

Exercise \(\PageIndex{11}\)

\(f(x)= \sqrt{x}+3\)

First, plot the graph of \(y = \sqrt{x}\), as shown in (a). Then, add 3 to produce the equation \(y = \sqrt{x} + 3\). This will shift the graph of of \(y = \sqrt{x}\) upward 3 units, as shown in (b).

Screen Shot 2019-05-28 at 11.50.22 PM.png

Project all points on the graph onto the x-axis to determine the domain: Domain = \([0, \infty)\). Project all points on the graph onto the y-axis to determine the range: Range = \([3, \infty)\).

Screen Shot 2019-05-29 at 10.35.29 AM.png

Exercise \(\PageIndex{12}\)

\(f(x)=\sqrt{x+3}\)

Exercise \(\PageIndex{13}\)

\(f(x)=\sqrt{x−2}\)

First, plot the graph of \(y = \sqrt{x}\), as shown in (a). Then, replace x with x − 2 to produce the equation \(y = \sqrt{x−2}\). This will shift the graph of \(y = \sqrt{x}\) to the right 2 units, as shown in (b).

Screen Shot 2019-05-29 at 10.20.45 AM.png

Project all points on the graph onto the x-axis to determine the domain: Domain = \([2, \infty)\). Project all points on the graph onto the y-axis to determine the range: Range = \([0, \infty)\).

Screen Shot 2019-05-29 at 10.35.43 AM.png

Exercise \(\PageIndex{14}\)

\(f(x)=\sqrt{x}−2\)

Exercise \(\PageIndex{15}\)

\(f(x)= \sqrt{x+5}+1\)

First, plot the graph of \(y = \sqrt{x}\), as shown in (a). Then, replace x with x + 5 to produce the equation \(y = \sqrt{x+5}\). Then add 1 to produce the equation \(f(x)= \sqrt{x+5}+1\). This will shift the graph of \(y = \sqrt{x}\) to the left 5 units, then upwards 1 unit, as shown in (b).

Screen Shot 2019-05-29 at 10.26.37 AM.png

Project all points on the graph onto the x-axis to determine the domain: Domain = \([−5, \infty)\). Project all points on the graph onto the y-axis to determine the range: Range = \([1, \infty)\).

Screen Shot 2019-05-29 at 10.36.25 AM.png

Exercise \(\PageIndex{16}\)

\(f(x)=\sqrt{x−2}−1\)

Exercise \(\PageIndex{17}\)

\(y = −\sqrt{x + 4}\)

First, plot the graph of \(y = \sqrt{x}\), as shown in (a). Then, negate to produce the \(y = −\sqrt{x}\). This will reflect the graph of \(y = \sqrt{x}\) across the x-axis as shown in (b). Finally, replace x with x + 4 to produce the equation \(y = −\sqrt{x + 4}\). This will shift the graph of \(y = −\sqrt{x}\) four units to the left, as shown in (c).

Screen Shot 2019-05-29 at 10.41.03 AM.png

Project all points on the graph onto the x-axis to determine the domain: Domain = \([−4, \infty)\). Project all points on the graph onto the y-axis to determine the range: Range = \((−\infty, 0]\).

Screen Shot 2019-05-29 at 10.36.47 AM.png

Exercise \(\PageIndex{18}\)

\(f(x)=−\sqrt{x}+4\)

Exercise \(\PageIndex{19}\)

\(f(x)=−\sqrt{x}+3\)

First, plot the graph of \(y = \sqrt{x}\), as shown in (a). Then, negate to produce the \(y = −\sqrt{x}\). This will reflect the graph of \(y = \sqrt{x}\) across the x-axis as shown in (b). Finally, add 3 to produce the equation \(y=−\sqrt{x}+3\). This will shift the graph of \(y = −\sqrt{x}\) three units upward, as shown in (c).

Screen Shot 2019-05-29 at 11.35.50 AM.png

Project all points on the graph onto the x-axis to determine the domain: Domain = \([0, \infty)\). Project all points on the graph onto the y-axis to determine the range: Range = \((−\infty, 3]\).

Screen Shot 2019-05-29 at 11.39.11 AM.png

Exercise \(\PageIndex{20}\)

Exercise \(\pageindex{21}\).

To draw the graph of the function \(f(x) = \sqrt{3−x}\), perform each of the following steps in sequence without the aid of a calculator.

  • Set up a coordinate system and sketch the graph of \(y = \sqrt{x}\). Label the graph with its equation.
  • Set up a second coordinate system and sketch the graph of \(y = \sqrt{−x}\). Label the graph with its equation.
  • Set up a third coordinate system and sketch the graph of \(y =\sqrt{−(x − 3)}\). Label the graph with its equation. This is the graph of \(y =\sqrt{3−x}\). Use interval notation to state the domain and range of this function.

First, plot the graph of \(y = \sqrt{x}\), as shown in (a). Then, replace x with − x to produce the equation \(y = \sqrt{−x}\). This will reflect the graph of \(y = \sqrt{x}\) across the y -axis, as shown in (b). Finally, replace x with x − 3 to produce the equation \( y = \sqrt{ − ( x − 3)}\). This will shift the graph of \(y = \sqrt{−x}\) three units to the right, as shown in (c).

Screen Shot 2019-05-29 at 11.51.00 AM.png

Project all points on the graph onto the x-axis to determine the domain: Domain = \((−\infty, 3]\). Project all points on the graph onto the y-axis to determine the range: Range = \([0, \infty)\).

Screen Shot 2019-05-29 at 11.52.37 AM.png

Exercise \(\PageIndex{22}\)

To draw the graph of the function \(f(x) = \sqrt{−x−3}\), perform each of the following steps in sequence.

  • Set up a third coordinate system and sketch the graph of \(y =\sqrt{−(x + 3)}\). Label the graph with its equation. This is the graph of \(y =\sqrt{−x−3}\). Use interval notation to state the domain and range of this function.

Exercise \(\PageIndex{23}\)

To draw the graph of the function \(f(x) = \sqrt{−x−3}\), perform each of the following steps in sequence without the aid of a calculator.

  • Set up a third coordinate system and sketch the graph of \(y =\sqrt{−(x + 1)}\). Label the graph with its equation. This is the graph of \(y =\sqrt{−x−1}\). Use interval notation to state the domain and range of this function.

First, plot the graph of \(y = \sqrt{x}\), as shown in (a). Then, replace x with −x to produce the equation \(y = \sqrt{−x}\). This will reflect the graph of \(y = \sqrt{x}\) across the y-axis, as shown in (b). Finally, replace x with x + 1 to produce the equation \(y = \sqrt{−(x + 1)}\). This will shift the graph of \(y = \sqrt{−x}\) one unit to the left, as shown in (c).

Screen Shot 2019-05-29 at 2.19.49 PM.png

Project all points on the graph onto the x-axis to determine the domain: Domain = \((−\infty, −1]\). Project all points on the graph onto the y-axis to determine the range: Range = \([0, \infty)\).

Screen Shot 2019-05-29 at 2.20.55 PM.png

Exercise \(\PageIndex{24}\)

To draw the graph of the function \(f(x) = \sqrt{1−x}\), perform each of the following steps in sequence.

  • Set up a third coordinate system and sketch the graph of \(y =\sqrt{−(x−1)}\). Label the graph with its equation. This is the graph of \(y =\sqrt{1−x}\). Use interval notation to state the domain and range of this function.

In Exercises 25 - 28 , perform each of the following tasks.

  • Draw the graph of the given function with your graphing calculator. Copy the image in your viewing window onto your homework paper. Label and scale each axis with xmin, xmax, ymin, and ymax. Label your graph with its equation. Use the graph to determine the domain of the function and describe the domain with interval notation.
  • Use a purely algebraic approach to determine the domain of the given function. Use interval notation to de- scribe your result. Does it agree with the graphical result from part 1?

Exercise \(\PageIndex{25}\)

\(f(x)= \sqrt{2x+7}\)

We use a graphing calculator to produce the following graph of \(f(x)= \sqrt{2x+7}\)

Screen Shot 2019-05-29 at 2.25.51 PM.png

We estimate that the domain will consist of all real numbers to the right of approximately − 3 . 5. To find an algebraic solution, note that you cannot take the square root of a negative number. Hence, the expression under the radical in \(f(x)= \sqrt{2x+7}\) must be greater than or equal to zero.

\(2x + 7 \ge 0\)

\(2x \ge −7\)

\(x \ge −\frac{7}{2}\)

Hence, the domain is \([−\frac{7}{2}, \infty)\).

Exercise \(\PageIndex{26}\)

\(f(x)= \sqrt{7−2x}\)

Exercise \(\PageIndex{27}\)

\(f(x)= \sqrt{12−4x}\)

We use a graphing calculator to produce the following graph of \(f(x)= \sqrt{12−4x}\).

Screen Shot 2019-05-29 at 2.31.07 PM.png

We estimate that the domain will consist of all real numbers to the right of approximately 3. To find an algebraic solution, note that you cannot take the square root of a negative number. Hence, the expression under the radical in \(f(x)= \sqrt{12−4x}\) must be greater than or equal to zero.

\(12−4x \ge 0\)

\(−4x \ge −12\)

\(x \le 3\)

Hence, the domain is \((−\infty, 3]\).

Exercise \(\PageIndex{28}\)

\(f(x)= \sqrt{12+2x}\)

In Exercises 29 - 40 , find the domain of the given function algebraically.

Exercise \(\PageIndex{29}\)

\(f(x)= \sqrt{2x+9}\)

The even root of a negative number is not defined as a real number. Thus, 2x + 9 must be greater than or equal to zero. Since \(2x + 9 \ge 0\) implies that \(x \ge −\frac{9}{2}\), the domain is the interval \([−\frac{9}{2},\infty)\).

Exercise \(\PageIndex{30}\)

\(f(x)=\sqrt{−3x+3}\)

Exercise \(\PageIndex{31}\)

\(f(x)=\sqrt{−8x−3}\)

The even root of a negative number is not defined as a real number. Thus, −8x−3 must be greater than or equal to zero. Since \(−8x−3 \ge 0\) implies that \(x \le −\frac{3}{8}\), the domain is the interval \((−\infty, −\frac{3}{8}]\).

Exercise \(\PageIndex{32}\)

\(f(x)=\sqrt{−3x+6}\)

Exercise \(\PageIndex{33}\)

\(f(x)=\sqrt{−6x−8}\)

The even root of a negative number is not defined as a real number. Thus, −6x−8 must be greater than or equal to zero. Since \(−6x−8 \ge 0\) implies that \(x \le −\frac{4}{3}\), the domain is the interval \((−\infty, \frac{4}{3}]\).

Exercise \(\PageIndex{34}\)

\(f(x)=\sqrt{8x−6}\)

Exercise \(\PageIndex{35}\)

\(f(x)=\sqrt{−7x+2}\)

The even root of a negative number is not defined as a real number. Thus, −7x+2 must be greater than or equal to zero. Since \(−7x+2 \ge 0\) implies that \(x \le \frac{2}{7}\), the domain is the interval \((−\infty, \frac{2}{7}]\).

Exercise \(\PageIndex{36}\)

\(f(x)=\sqrt{8x−3}\)

Exercise \(\PageIndex{37}\)

\(f(x)=\sqrt{6x+3}\)

The even root of a negative number is not defined as a real number. Thus, 6x+3 must be greater than or equal to zero. Since \(6x+3 \ge 0\) implies that \(x \ge −\frac{1}{2}\), the domain is the interval \([−\frac{1}{2}, \infty)\).

Exercise \(\PageIndex{38}\)

\(f(x)=\sqrt{x−5}\)

Exercise \(\PageIndex{39}\)

\(f(x)=\sqrt{−7x−8}\)

The even root of a negative number is not defined as a real number. Thus, −7x−8 must be greater than or equal to zero. Since \(−7x−8 \ge 0\) implies that \(x \le −\frac{8}{7}\), the domain is the interval \((−\infty, −\frac{8}{7}]\)

Exercise \(\PageIndex{40}\)

\(f(x)=\sqrt{7x+8}\)

  • NCERT Solutions
  • NCERT Class 8
  • NCERT 8 Maths
  • Chapter 6: Square Square Roots

NCERT Solutions for Class 8 Maths Chapter 6 Squares and Square Roots

Ncert solutions class 8 maths chapter 6 – free pdf download.

NCERT Solutions for Class 8 Maths Chapter 6 Squares and Square Roots are beneficial for students since it aids them in scoring high marks in the exam. The subject experts at BYJU’S outline the concepts in a distinct and well-defined manner, keeping the IQ level of students in mind. These solution modules use various shortcut hints and practical examples to explain all the exercise questions in simple and easily understandable language. Solving the NCERT Class 8 Solutions is a must to obtain an excellent score in the examination.

Download Exclusively Curated Chapter Notes for Class 8 Maths Chapter – 6 Squares and Square Roots

Download most important questions for class 8 maths chapter – 6 squares and square roots.

Here, with the NCERT Solutions provided, the students will learn various techniques to determine whether a given natural number is a perfect square or not. The answers to all problems within this chapter present in NCERT Textbooks are provided here in a detailed and step-by-step way to help the students understand more effectively. Students can download the NCERT Solutions for Class 8 Maths Chapter 6 from the link provided below.

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NCERT Solutions for Class 8 Maths March 28 Chapter 6 Squares and Square Roots

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Access Answers of NCERT Class 8 Maths Chapter 6 – Squares and Square Roots

Exercise 6.1 Page: 96

1. What will be the unit digit of the squares of the following numbers?

viii. 99880

The unit digit of square of a number having ‘a’ at its unit place ends with a×a.

i. The unit digit of the square of a number having digit 1 as unit’s place is 1.

∴ Unit digit of the square of number 81 is equal to 1.

ii. The unit digit of the square of a number having digit 2 as unit’s place is 4.

∴ Unit digit of the square of number 272 is equal to 4.

iii. The unit digit of the square of a number having digit 9 as unit’s place is 1.

∴ Unit digit of the square of number 799 is equal to 1.

iv. The unit digit of the square of a number having digit 3 as unit’s place is 9.

∴ Unit digit of the square of number 3853 is equal to 9.

v. The unit digit of the square of a number having digit 4 as unit’s place is 6.

∴ Unit digit of the square of number 1234 is equal to 6.

vi. The unit digit of the square of a number having digit 7 as unit’s place is 9.

∴ Unit digit of the square of number 26387 is equal to 9.

vii. The unit digit of the square of a number having digit 8 as unit’s place is 4.

∴ Unit digit of the square of number 52698 is equal to 4.

square root question with solution

viii. The unit digit of the square of a number having digit 0 as unit’s place is 01.

∴ Unit digit of the square of number 99880 is equal to 0.

ix. The unit digit of the square of a number having digit 6 as unit’s place is 6.

∴ Unit digit of the square of number 12796 is equal to 6.

x. The unit digit of the square of a number having digit 5 as unit’s place is 5.

∴ Unit digit of the square of number 55555 is equal to 5.

2. The following numbers are obviously not perfect squares. Give reason.

vii. 222000

viii. 505050

We know that natural numbers ending in the digits 0, 2, 3, 7 and 8 are not perfect squares.

i. 1057 ⟹ Ends with 7

ii. 23453 ⟹ Ends with 3

iii. 7928 ⟹ Ends with 8

iv. 222222 ⟹ Ends with 2

square root question with solution

v. 64000 ⟹ Ends with 0

vi. 89722 ⟹ Ends with 2

vii. 222000 ⟹ Ends with 0

viii. 505050 ⟹ Ends with 0

3. The squares of which of the following would be odd numbers?

We know that the square of an odd number is odd and the square of an even number is even.

i. The square of 431 is an odd number.

ii. The square of 2826 is an even number.

iii. The square of 7779 is an odd number.

iv. The square of 82004 is an even number.

4. Observe the following pattern and find the missing numbers. 11 2 = 121

101 2 = 10201

1001 2 = 1002001

100001 2 = 1 …….2………1

10000001 2 = ……………………..

We observe that the square on the number on R.H.S of the equality has an odd number of digits such that the first and last digits both are 1 and middle digit is 2. And the number of zeros between left most digits 1 and the middle digit 2 and right most digit 1 and the middle digit 2 is same as the number of zeros in the given number.

∴ 100001 2 = 10000200001

10000001 2 = 100000020000001

5. Observe the following pattern and supply the missing numbers. 112 = 121

1012 = 10201

101012 = 102030201

10101012 = ………………………

…………2 = 10203040504030201

We observe that the square on the number on R.H.S of the equality has an odd number of digits such that the first and last digits both are 1. And, the square is symmetric about the middle digit. If the middle digit is 4, then the number to be squared is 10101 and its square is 102030201.

So, 10101012 =1020304030201

1010101012 =10203040505030201

6. Using the given pattern, find the missing numbers. 1 2 + 2 2 + 2 2 = 3 2

2 2 + 3 2 + 6 2 = 7 2

3 2 + 4 2 + 12 2 = 13 2

4 2 + 5 2 + _2 = 21 2

5 + _ 2 + 30 2 = 31 2

6 + 7 + _ 2 = _ 2

Given, 1 2 + 2 2 + 2 2 = 3 2

i.e 1 2 + 2 2 + (1×2 ) 2 = ( 1 2 + 2 2 -1 × 2 ) 2

2 2 + 3 2 + 6 2 =7 2

∴ 2 2 + 3 2 + (2×3 ) 2 = (2 2 + 3 2 -2 × 3) 2

∴ 3 2 + 4 2 + (3×4 ) 2 = (3 2 + 4 2 – 3 × 4) 2

4 2 + 5 2 + (4×5 ) 2 = (4 2 + 5 2 – 4 × 5) 2

∴ 4 2 + 5 2 + 20 2 = 21 2

5 2 + 6 2 + (5×6 ) 2 = (5 2 + 6 2 – 5 × 6) 2

∴ 5 2 + 6 2 + 30 2 = 31 2

6 2 + 7 2 + (6×7 ) 2 = (6 2 + 7 2 – 6 × 7) 2

∴ 6 2 + 7 2 + 42 2 = 43 2

7. Without adding, find the sum.

i. 1 + 3 + 5 + 7 + 9

Sum of first five odd number = (5) 2 = 25

ii. 1 + 3 + 5 + 7 + 9 + I1 + 13 + 15 + 17 +19

Sum of first ten odd number = (10) 2 = 100

iii. 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23

Sum of first thirteen odd number = (12) 2 = 144

square root question with solution

8. (i) Express 49 as the sum of 7 odd numbers.

We know, sum of first n odd natural numbers is n 2 . Since,49 = 7 2

∴ 49 = sum of first 7 odd natural numbers = 1 + 3 + 5 + 7 + 9 + 11 + 13

(ii) Express 121 as the sum of 11 odd numbers.

Since, 121 = 11 2

∴ 121 = sum of first 11 odd natural numbers = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21

square root question with solution

9. How many numbers lie between squares of the following numbers?

i. 12 and 13

ii. 25 and 26

iii. 99 and 100

Between n 2 and (n+1) 2 , there are 2n non–perfect square numbers.

i. 122 and 132 there are 2×12 = 24 natural numbers.

ii. 252 and 262 there are 2×25 = 50 natural numbers.

iii. 992 and 1002 there are 2×99 =198 natural numbers.

square root question with solution

Exercise 6.2 Page: 98

1. Find the square of the following numbers.

= (30 +2) 2

= (30) 2 + (2) 2 + 2×30×2 [Since, (a+b) 2 = a 2 +b 2 +2ab]

= 900 + 4 + 120

square root question with solution

= (30+5 ) 2

= (30) 2 + (5) 2 + 2×30×5 [Since, (a+b) 2 = a 2 +b 2 +2ab]

= 900 + 25 + 300

iii. (86) 2

= (90 – 4) 2

= (90) 2 + (4) 2 – 2×90×4 [Since, (a+b) 2 = a 2 +b 2 +2ab]

= 8100 + 16 – 720

= 8116 – 720

= (90+3 ) 2

= (90) 2 + (3) 2 + 2×90×3 [Since, (a+b) 2 = a 2 +b 2 +2ab]

= 8100 + 9 + 540

square root question with solution

= (70+1 ) 2

= (70) 2 + (1) 2 +2×70×1 [Since, (a+b) 2 = a 2 +b 2 +2ab]

= 4900 + 1 + 140

square root question with solution

= (50 -4 ) 2

= (50) 2 + (4) 2 – 2×50×4 [Since, (a+b) 2 = a 2 +b 2 +2ab]

= 2500 + 16 – 400

2. Write a Pythagorean triplet whose one member is.

For any natural number m, we know that 2m, m2–1, m2+1 is a Pythagorean triplet.

⇒ m = 6/2 = 3

m2–1= 32 – 1 = 9–1 = 8

m2+1= 32+1 = 9+1 = 10

∴ (6, 8, 10) is a Pythagorean triplet.

square root question with solution

ii. 2m = 14

⇒ m = 14/2 = 7

m2–1= 72–1 = 49–1 = 48

m2+1 = 72+1 = 49+1 = 50

∴ (14, 48, 50) is not a Pythagorean triplet.

square root question with solution

iii. 2m = 16

⇒ m = 16/2 = 8

m2–1 = 82–1 = 64–1 = 63

m2+ 1 = 82+1 = 64+1 = 65

∴ (16, 63, 65) is a Pythagorean triplet.

iv. 2m = 18

⇒ m = 18/2 = 9

m2–1 = 92–1 = 81–1 = 80

m2+1 = 92+1 = 81+1 = 82

∴ (18, 80, 82) is a Pythagorean triplet.

square root question with solution

Exercise 6.3 Page: 102

1. What could be the possible ‘one’s’ digits of the square root of each of the following numbers?

iii. 998001

iv. 657666025

i. We know that the unit’s digit of the square of a number having digit as unit’s

place 1 is 1 and also 9 is 1[9 2 =81 whose unit place is 1].

∴ Unit’s digit of the square root of number 9801 is equal to 1 or 9.

ii. We know that the unit’s digit of the square of a number having digit as unit’s

place 6 is 6 and also 4 is 6 [6 2 =36 and 4 2 =16, both the squares have unit digit 6].

∴ Unit’s digit of the square root of number 99856 is equal to 6.

iii. We know that the unit’s digit of the square of a number having digit as unit’s

∴ Unit’s digit of the square root of number 998001 is equal to 1 or 9.

iv. We know that the unit’s digit of the square of a number having digit as unit’s

place 5 is 5.

∴ Unit’s digit of the square root of number 657666025 is equal to 5.

2. Without doing any calculation, find the numbers which are surely not perfect squares.

We know that natural numbers ending with the digits 0, 2, 3, 7 and 8 are not perfect square.

i. 153⟹ Ends with 3.

∴, 153 is not a perfect square

ii. 257⟹ Ends with 7

∴, 257 is not a perfect square

iii. 408⟹ Ends with 8

∴, 408 is not a perfect square

iv. 441⟹ Ends with 1

∴, 441 is a perfect square.

3. Find the square roots of 100 and 169 by the method of repeated subtraction .

100 – 1 = 99

99 – 3 = 96

96 – 5 = 91

91 – 7 = 84

84 – 9 = 75

75 – 11 = 64

64 – 13 = 51

51 – 15 = 36

36 – 17 = 19

19 – 19 = 0

Here, we have performed subtraction ten times.

∴ √100 = 10

169 – 1 = 168

168 – 3 = 165

165 – 5 = 160

160 – 7 = 153

153 – 9 = 144

144 – 11 = 133

133 – 13 = 120

120 – 15 = 105

105 – 17 = 88

88 – 19 = 69

69 – 21 = 48

48 – 23 = 25

25 – 25 = 0

Here, we have performed subtraction thirteen times.

∴ √169 = 13

square root question with solution

4. Find the square roots of the following numbers by the Prime Factorisation Method.

NCERT Solution For Class 8 Maths Chapter 6 Image 1

729 = 3×3×3×3×3×3×1

⇒ 729 = (3×3)×(3×3)×(3×3)

⇒ 729 = (3×3×3)×(3×3×3)

⇒ 729 = (3×3×3) 2

⇒ √729 = 3×3×3 = 27

NCERT Solution For Class 8 Maths Chapter 6 Image 2

400 = 2×2×2×2×5×5×1

⇒ 400 = (2×2)×(2×2)×(5×5)

⇒ 400 = (2×2×5)×(2×2×5)

⇒ 400 = (2×2×5) 2

⇒ √400 = 2×2×5 = 20

NCERT Solution For Class 8 Maths Chapter 6 Image 3

1764 = 2×2×3×3×7×7

⇒ 1764 = (2×2)×(3×3)×(7×7)

⇒ 1764 = (2×3×7)×(2×3×7)

⇒ 1764 = (2×3×7) 2

⇒ √1764 = 2 ×3×7 = 42

square root question with solution

4096 = 2×2×2×2×2×2×2×2×2×2×2×2

⇒ 4096 = (2×2)×(2×2)×(2×2)×(2×2)×(2×2)×(2×2)

⇒ 4096 = (2×2×2×2×2×2)×(2×2×2×2×2×2)

⇒ 4096 = (2×2×2×2×2×2) 2

⇒ √4096 = 2×2×2 ×2×2×2 = 64

NCERT Solution For Class 8 Maths Chapter 6 Image 5

7744 = 2×2×2×2×2×2×11×11×1

⇒ 7744 = (2×2)×(2×2)×(2×2)×(11×11)

⇒ 7744 = (2×2×2×11)×(2×2×2×11)

⇒ 7744 = (2×2×2×11) 2

⇒ √7744 = 2×2×2×11 = 88

NCERT Solution For Class 8 Maths Chapter 6 Image 6

9604 = 62 × 2 × 7 × 7 × 7 × 7

⇒ 9604 = ( 2 × 2 ) × ( 7 × 7 ) × ( 7 × 7 )

⇒ 9604 = ( 2 × 7 ×7 ) × ( 2 × 7 ×7 )

⇒ 9604 = ( 2×7×7 ) 2

⇒ √9604 = 2×7×7 = 98

NCERT Solution For Class 8 Maths Chapter 6 Image 7

5929 = 7×7×11×11

⇒ 5929 = (7×7)×(11×11)

⇒ 5929 = (7×11)×(7×11)

⇒ 5929 = (7×11) 2

⇒ √5929 = 7×11 = 77

NCERT Solution For Class 8 Maths Chapter 6 Image 7

9216 = 2×2×2×2×2×2×2×2×2×2×3×3×1

⇒ 9216 = (2×2)×(2×2) × ( 2 × 2 ) × ( 2 × 2 ) × ( 2 × 2 ) × ( 3 × 3 )

⇒ 9216 = ( 2 × 2 × 2 × 2 × 2 × 3) × ( 2 × 2 × 2 × 2 × 2 × 3)

⇒ 9216 = 96 × 96

⇒ 9216 = ( 96 ) 2

⇒ √9216 = 96

NCERT Solution For Class 8 Maths Chapter 6 Image 8

529 = 23×23

529 = (23) 2

NCERT Solution For Class 8 Maths Chapter 6 Image 9

8100 = 2×2×3×3×3×3×5×5×1

⇒ 8100 = (2×2) ×(3×3)×(3×3)×(5×5)

⇒ 8100 = (2×3×3×5)×(2×3×3×5)

⇒ 8100 = 90×90

⇒ 8100 = (90) 2

⇒ √8100 = 90

5. For each of the following numbers, find the smallest whole number by which it should be multiplied so as to get a perfect square number. Also find the square root of the square number so obtained.

NCERT Solution For Class 8 Maths Chapter 6 Image 10

252 = 2×2×3×3×7

= (2×2)×(3×3)×7

Here, 7 cannot be paired.

∴ We will multiply 252 by 7 to get perfect square.

New number = 252 × 7 = 1764

NCERT Solution For Class 8 Maths Chapter 6 Image 11

⇒ 1764 = 2 2 ×3 2 ×7 2

⇒ √1764 = 2×3×7 = 42

NCERT Solution For Class 8 Maths Chapter 6 Image 12

180 = 2×2×3×3×5

= (2×2)×(3×3)×5

Here, 5 cannot be paired.

∴ We will multiply 180 by 5 to get perfect square.

New number = 180 × 5 = 900

NCERT Solution For Class 8 Maths Chapter 6 Image 13

900 = 2×2×3×3×5×5×1

⇒ 900 = (2×2)×(3×3)×(5×5)

⇒ 900 = 2 2 ×3 2 ×5 2

⇒ 900 = (2×3×5) 2

⇒ √900 = 2×3×5 = 30

NCERT Solution For Class 8 Maths Chapter 6 Image 14

1008 = 2×2×2×2×3×3×7

= (2×2)×(2×2)×(3×3)×7

∴ We will multiply 1008 by 7 to get perfect square.

New number = 1008×7 = 7056

NCERT Solution For Class 8 Maths Chapter 6 Image 15

7056 = 2×2×2×2×3×3×7×7

⇒ 7056 = (2×2)×(2×2)×(3×3)×(7×7)

⇒ 7056 = 2 2 ×2 2 ×3 2 ×7 2

⇒ 7056 = (2×2×3×7) 2

⇒ √7056 = 2×2×3×7 = 84

NCERT Solution For Class 8 Maths Chapter 6 Image 16

2028 = 2×2×3×13×13

= (2×2)×(13×13)×3

Here, 3 cannot be paired.

∴ We will multiply 2028 by 3 to get perfect square. New number = 2028×3 = 6084

NCERT Solution For Class 8 Maths Chapter 6 Image 17

6084 = 2×2×3×3×13×13

⇒ 6084 = (2×2)×(3×3)×(13×13)

⇒ 6084 = 2 2 ×3 2 ×13 2

⇒ 6084 = (2×3×13) 2

⇒ √6084 = 2×3×13 = 78

NCERT Solution For Class 8 Maths Chapter 6 Image 18

1458 = 2×3×3×3×3×3×3

= (3×3)×(3×3)×(3×3)×2

Here, 2 cannot be paired.

∴ We will multiply 1458 by 2 to get perfect square. New number = 1458 × 2 = 2916

NCERT Solution For Class 8 Maths Chapter 6 Image 19

2916 = 2×2×3×3×3×3×3×3

⇒ 2916 = (3×3)×(3×3)×(3×3)×(2×2)

⇒ 2916 = 3 2 ×3 2 ×3 2 ×2 2

⇒ 2916 = (3×3×3×2) 2

⇒ √2916 = 3×3×3×2 = 54

NCERT Solution For Class 8 Maths Chapter 6 Image 20

768 = 2×2×2×2×2×2×2×2×3

= (2×2)×(2×2)×(2×2)×(2×2)×3

∴ We will multiply 768 by 3 to get perfect square.

New number = 768×3 = 2304

NCERT Solution For Class 8 Maths Chapter 6 Image 21

2304 = 2×2×2×2×2×2×2×2×3×3

⇒ 2304 = (2×2)×(2×2)×(2×2)×(2×2)×(3×3)

⇒ 2304 = 2 2 ×2 2 ×2 2 ×2 2 ×3 2

⇒ 2304 = (2×2×2×2×3) 2

⇒ √2304 = 2×2×2×2×3 = 48

6. For each of the following numbers, find the smallest whole number by which it should be divided so as to get a perfect square. Also find the square root of the square number so obtained.

NCERT Solution For Class 8 Maths Chapter 6 Image 22

∴ We will divide 252 by 7 to get perfect square. New number = 252 ÷ 7 = 36

NCERT Solution For Class 8 Maths Chapter 6 Image 23

36 = 2×2×3×3

⇒ 36 = (2×2)×(3×3)

⇒ 36 = 2 2 ×3 2

⇒ 36 = (2×3) 2

⇒ √36 = 2×3 = 6

NCERT Solution For Class 8 Maths Chapter 6 Image 24

2925 = 3×3×5×5×13

= (3×3)×(5×5)×13

Here, 13 cannot be paired.

∴ We will divide 2925 by 13 to get perfect square. New number = 2925 ÷ 13 = 225

NCERT Solution For Class 8 Maths Chapter 6 Image 24

225 = 3×3×5×5

⇒ 225 = (3×3)×(5×5)

⇒ 225 = 3 2 ×5 2

⇒ 225 = (3×5) 2

⇒ √36 = 3×5 = 15

NCERT Solution For Class 8 Maths Chapter 6 Image 25

396 = 2×2×3×3×11

= (2×2)×(3×3)×11

Here, 11 cannot be paired.

∴ We will divide 396 by 11 to get perfect square. New number = 396 ÷ 11 = 36

NCERT Solution For Class 8 Maths Chapter 6 Image 26

2645 = 5×23×23

⇒ 2645 = (23×23)×5

∴ We will divide 2645 by 5 to get perfect square.

New number = 2645 ÷ 5 = 529

NCERT Solution For Class 8 Maths Chapter 6 Image 28

⇒ 529 = (23) 2

⇒ √529 = 23

NCERT Solution For Class 8 Maths Chapter 6 Image 29

2800 = 2×2×2×2×5×5×7

= (2×2)×(2×2)×(5×5)×7

∴ We will divide 2800 by 7 to get perfect square. New number = 2800 ÷ 7 = 400

NCERT Solution For Class 8 Maths Chapter 6 Image 30

400 = 2×2×2×2×5×5

⇒ √400 = 20

NCERT Solution For Class 8 Maths Chapter 6 Image 31

1620 = 2×2×3×3×3×3×5

= (2×2)×(3×3)×(3×3)×5

∴ We will divide 1620 by 5 to get perfect square. New number = 1620 ÷ 5 = 324

NCERT Solution For Class 8 Maths Chapter 6 Image 32

324 = 2×2×3×3×3×3

⇒ 324 = (2×2)×(3×3)×(3×3)

⇒ 324 = (2×3×3) 2

⇒ √324 = 18

7. The students of Class VIII of a school donated Rs 2401 in all, for Prime Minister’s National Relief Fund. Each student donated as many rupees as the number of students in the class. Find the number of students in the class.

Let the number of students in the school be, x.

∴ Each student donate Rs.x .

Total amount contributed by all the students= x×x=x 2 Given, x 2 = Rs.2401

NCERT Solution For Class 8 Maths Chapter 6 Image 33

x 2 = 7×7×7×7

⇒ x 2 = (7×7)×(7×7)

⇒ x 2 = 49×49

⇒ x = √(49×49)

∴ The number of students = 49

square root question with solution

8. 2025 plants are to be planted in a garden in such a way that each row contains as many plants as the number of rows. Find the number of rows and the number of plants in each row.

Let the number of rows be, x.

∴ the number of plants in each rows = x.

Total plants to be planted in the garden = x × x =x 2

x 2 = Rs.2025

NCERT Solution For Class 8 Maths Chapter 6 Image 34

x 2 = 3×3×3×3×5×5

⇒ x 2 = (3×3)×(3×3)×(5×5)

⇒ x2 = (3×3×5)×(3×3×5)

⇒ x2 = 45×45

⇒ x = √45×45

∴ The number of rows = 45 and the number of plants in each rows = 45.

9. Find the smallest square number that is divisible by each of the numbers 4, 9 and 10.

NCERT Solution For Class 8 Maths Chapter 6 Image 35

L.C.M of 4, 9 and 10 is (2×2×9×5) 180.

180 = 2×2×9×5

= (2×2)×3×3×5

∴ we will multiply 180 by 5 to get perfect square.

Hence, the smallest square number divisible by 4, 9 and 10 = 180×5 = 900

10. Find the smallest square number that is divisible by each of the numbers 8, 15 and 20.

NCERT Solution For Class 8 Maths Chapter 6 Image 36

L.C.M of 8, 15 and 20 is (2×2×5×2×3) 120.

120 = 2×2×3×5×2

= (2×2)×3×5×2

Here, 3, 5 and 2 cannot be paired.

∴ We will multiply 120 by (3×5×2) 30 to get perfect square.

Hence, the smallest square number divisible by 8, 15 and 20 =120×30 = 3600

Exercise 6.4 Page: 107

1. Find the square root of each of the following numbers by Division method.

NCERT Solution For Class 8 Maths Chapter 6 Image 37

∴ √2304 = 48

NCERT Solution For Class 8 Maths Chapter 6 Image 38

∴ √4489 = 67

NCERT Solution For Class 8 Maths Chapter 6 Image 39

∴ √3481 = 59

NCERT Solution For Class 8 Maths Chapter 6 Image 40

∴ √529 = 23

NCERT Solution For Class 8 Maths Chapter 6 Image 41

∴ √3249 = 57

NCERT Solution For Class 8 Maths Chapter 6 Image 42

∴ √1369 = 37

NCERT Solution For Class 8 Maths Chapter 6 Image 43

∴ √5776 = 76

NCERT Solution For Class 8 Maths Chapter 6 Image 44

∴ √7921 = 89

NCERT Solution For Class 8 Maths Chapter 6 Image 45

∴ √576 = 24

NCERT Solution For Class 8 Maths Chapter 6 Image 46

∴ √1024 = 32

NCERT Solution For Class 8 Maths Chapter 6 Image 47

∴ √3136 = 56

NCERT Solution For Class 8 Maths Chapter 6 Image 48

∴ √900 = 30

2. Find the number of digits in the square root of each of the following numbers (without any

calculation).64

NCERT Solution For Class 8 Maths Chapter 6 Image 49

∴ √144 = 12

Hence, the square root of the number 144 has 2 digits.

NCERT Solution For Class 8 Maths Chapter 6 Image 50

Hence, the square root of the number 4489 has 2 digits.

NCERT Solution For Class 8 Maths Chapter 6 Image 51

√27225 = 165

Hence, the square root of the number 27225 has 3 digits.

NCERT Solution For Class 8 Maths Chapter 6 Image 52

∴ √390625 = 625

Hence, the square root of the number 390625 has 3 digits.

3. Find the square root of the following decimal numbers.

NCERT Solution For Class 8 Maths Chapter 6 Image 53

∴ √2.56 = 1.6

square root question with solution

∴ √7.29 = 2.7

square root question with solution

∴ √51.84 = 7.2

square root question with solution

∴ √42.25 = 6.5

NCERT Solution For Class 8 Maths Chapter 6 Image 57

∴ √31.36 = 5.6

square root question with solution

4. Find the least number which must be subtracted from each of the following numbers so as to get a perfect square. Also find the square root of the perfect square so obtained.

NCERT Solution For Class 8 Maths Chapter 6 Image 58

∴ We must subtract 2 from 402 to get a perfect square.

New number = 402 – 2 = 400

NCERT Solution For Class 8 Maths Chapter 6 Image 59

∴ √400 = 20

NCERT Solution For Class 8 Maths Chapter 6 Image 60

∴ We must subtract 53 from 1989 to get a perfect square. New number = 1989 – 53 = 1936

NCERT Solution For Class 8 Maths Chapter 6 Image 61

∴ √1936 = 44

NCERT Solution For Class 8 Maths Chapter 6 Image 62

∴ We must subtract 1 from 3250 to get a perfect square.

New number = 3250 – 1 = 3249

NCERT Solution For Class 8 Maths Chapter 6 Image 63

∴ We must subtract 41 from 825 to get a perfect square.

New number = 825 – 41 = 784

NCERT Solution For Class 8 Maths Chapter 6 Image 65

∴ √784 = 28

NCERT Solution For Class 8 Maths Chapter 6 Image 66

∴ We must subtract 31 from 4000 to get a perfect square. New number = 4000 – 31 = 3969

∴ √3969 = 63

5. Find the least number which must be added to each of the following numbers so as to get a perfect square. Also find the square root of the perfect square so obtained.

NCERT Solution For Class 8 Maths Chapter 6 Image 67

Here, (22)2 < 525 > (23)2

We can say 525 is ( 129 – 125 ) 4 less than (23)2.

∴ If we add 4 to 525, it will be perfect square. New number = 525 + 4 = 529

NCERT Solution For Class 8 Maths Chapter 6 Image 69

Here, (41)2 < 1750 > (42) 2

We can say 1750 is ( 164 – 150 ) 14 less than (42) 2 .

∴ If we add 14 to 1750, it will be perfect square.

New number = 1750 + 14 = 1764

NCERT Solution For Class 8 Maths Chapter 6 Image 72

∴√1764 = 42

NCERT Solution For Class 8 Maths Chapter 6 Image 73

Here, (15)2 < 252 > (16)2

We can say 252 is ( 156 – 152 ) 4 less than (16)2.

∴ If we add 4 to 252, it will be perfect square.

New number = 252 + 4 = 256

NCERT Solution For Class 8 Maths Chapter 6 Image 75

∴ √256 = 16

NCERT Solution For Class 8 Maths Chapter 6 Image 76

Here, (42)2 < 1825 > (43)2

We can say 1825 is ( 249 – 225 ) 24 less than (43)2.

∴ If we add 24 to 1825, it will be perfect square.

New number = 1825 + 24 = 1849

NCERT Solution For Class 8 Maths Chapter 6 Image 78

∴ √1849 = 43

NCERT Solution For Class 8 Maths Chapter 6 Image 79

Here, (80)2 < 6412 > (81)2

We can say 6412 is ( 161 – 12 ) 149 less than (81)2.

∴ If we add 149 to 6412, it will be perfect square.

New number = 6412 + 149 = 656

NCERT Solution For Class 8 Maths Chapter 6 Image 81

∴ √6561 = 81

6. Find the length of the side of a square whose area is 441 m2.

Let the length of each side of the field = a Then, area of the field = 441 m2

⇒ a2 = 441 m2

⇒a = √441 m

NCERT Solution For Class 8 Maths Chapter 6 Image 82

∴ The length of each side of the field = a m = 21 m.

7. In a right triangle ABC, ∠B = 90°.

a. If AB = 6 cm, BC = 8 cm, find AC

b. If AC = 13 cm, BC = 5 cm, find AB

NCERT Solution For Class 8 Maths Chapter 6 Image 83

Given, AB = 6 cm, BC = 8 cm

Let AC be x cm.

∴ AC2 = AB2 + BC2

NCERT Solution For Class 8 Maths Chapter 6 Image 84

Hence, AC = 10 cm.

NCERT Solution For Class 8 Maths Chapter 6 Image 85

Given, AC = 13 cm, BC = 5 cm

Let AB be x cm.

⇒ AC2 – BC2 = AB2

NCERT Solution For Class 8 Maths Chapter 6 Image 86

Hence, AB = 12 cm

8. A gardener has 1000 plants. He wants to plant these in such a way that the number of rows

and the number of columns remain same. Find the minimum number of plants he needs more for this.

Let the number of rows and column be, x.

∴ Total number of row and column= x× x = x2 As per question, x2 = 1000

⇒ x = √1000

NCERT Solution For Class 8 Maths Chapter 6 Image 87

Here, (31)2 < 1000 > (32)2

We can say 1000 is ( 124 – 100 ) 24 less than (32)2.

∴ 24 more plants are needed.

9. There are 500 children in a school. For a P.T. drill they have to stand in such a manner that the number of rows is equal to number of columns. How many children would be left out in this arrangement.

∴ Total number of row and column= x × x = x2 As per question, x2 = 500

NCERT Solution For Class 8 Maths Chapter 6 Image 88

Hence, 16 children would be left out in the arrangement

NCERT Solutions for Class 8 Maths Chapter 6 – Squares and Square Roots

The NCERT Solutions for Class 8 Maths Chapter 8 deals with the concept of squares and square roots, along with other main topics and concepts.

The major concepts covered in this chapter include:

6.1 Introduction

6.2 Properties of Square Numbers

6.3 Some interesting patterns

6.4 Finding the square of a number

6.4.1 Patterns in Squares

6.4.2 Pythagorean triplets

6.5 Square Roots

6.5.1 Finding square roots

6.5.2 Finding square root through repeated subtraction

6.5.3 Finding square root through prime factorisation

6.5.4 Finding square Root by division method

6.6 Square Roots of Decimals

6.7 Estimating Square Root

Below are the exercise-wise NCERT Solutions for Class 8 Maths Chapter 6:

Exercise 6.1 Solutions 9 Questions

Exercise 6.2 Solutions 2 Questions

Exercise 6.3 Solutions 10 Questions

Exercise 6.4 Solutions 9 Questions

NCERT Solutions for Class 8 Maths Chapter 6 Squares and Square Roots – Summary

Chapter 6 of NCERT Solutions for Class 8 Maths discusses the following:

  • If a natural number m can be expressed as n 2 , where n is a natural number, then m is a square number.
  • All square numbers end with 0, 1, 4, 5, 6 or 9 at units place.
  • Square numbers can only have an even number of zeros at the end.
  • Square root is the inverse operation of square.
  • There are two integral square roots of a perfect square number

Learning the chapter Squares and Square Roots enables the students to understand:

  • Square and Square roots
  • Square roots using the factor method and division method for numbers containing(a) no more than a total of 4 digits and (b) no more than 2 decimal places
  • Estimating square roots
  • Learning the process of moving nearer to the required number

Disclaimer:

Dropped Topics – 6.7 Estimating square roots.

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  1. Square and Square Roots Questions [Solved]

    Let us solve a few questions based on square and square roots. Question 1: Find the difference between the following without actual calculation: (i) 35 2 - 34 2 (ii) 134 2 - 133 2 (iii) 62 2 - 61 2. Solution: The difference between the squares of two consecutive natural numbers is the sum of the natural numbers. (i) 35 2 - 34 2 = 35 + 34 = 69

  2. 72 Square Root Questions with answers for practice

    Online Practice questions quiz on square and Square root (Quantitative Aptitude). Ques. Find square root of 324. (a) 18(b) 16 (c) 14 (d) 22

  3. Solving square-root equations (article)

    Practice question 1: Isolating the radical term Solve the following equation for x . 2 x − 7 − 3 = 4 x = [Show answer.] Practice question 2: Two possible solutions What are the solutions to the following equation? 6 + 3 x = 2 x + 12 + 2 x Choose all answers that apply: x = 0 A x = 0 x = − 4 B x = − 4 x = 4 C x = 4 x = − 6 D x = − 6 x = 1 E x = 1

  4. How to Solve Square Root Problems (with Pictures)

    1 Square a number by multiplying it by itself. To understand square roots, it's best to start with squares. Squares are easy—taking the square of a number is just multiplying it by itself. [1] For instance, 3 squared is the same as 3 × 3 = 9 and 9 squared is the same as 9 × 9 = 81.

  5. Solving quadratics by taking square roots

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  6. 8.6: Solve Equations with Square Roots

    OpenStax Learning Objectives By the end of this section, you will be able to: Solve radical equations Use square roots in applications Note Before you get started, take this readiness quiz.

  7. Intro to square-root equations & extraneous solutions

    6 years ago The original equation is (sqrt)x=2x-6. When you see a radical with no + or - sign before it, we assume that we are only taking the principal root (the positive version). When he plugs that 2.25 into the equation, he can't use -1.5 because of that. 3 comments ( 23 votes) Upvote

  8. 9.1 Solve Quadratic Equations Using the Square Root Property

    Solve Quadratic Equations of the form ax2 = k a x 2 = k using the Square Root Property. We have already solved some quadratic equations by factoring. Let's review how we used factoring to solve the quadratic equation x2 = 9. x2 = 9. x 2 = 9. x 2 = 9. Put the equation in standard form. x2 − 9 = 0. x 2 − 9 = 0.

  9. Square Root Questions

    Answer: Square numbers can be defined as the numbers that are produced when a number is multiplied by itself. For example, if n is a number and it is multiplied by itself, then the square of n can be given as n 2.Another example is the square of 10, which is 10 2 = 10 x 10 = 100.. Further, the square root of a number can be explained as the value which when multiplied by itself gives the ...

  10. 11.1: The Square Root Property

    Solution. We could rewrite the equation so that 81 81 is on the left and then solve by factoring. However, for the sake of the property, we solve this equation by applying the square root property. x2 = 81 x = ± 81−−√ x = ±9 x = 9 or x = −9 The x2 is isolated and we apply the square root property Simplify Rewrite as two solutions ...

  11. Free square root worksheets (PDF and html)

    The option "Only simplify, no answers as decimals" forces the answer NOT to be given as a rounded decimal, but instead the answer is simplified if possible, and the square root is left in the answer if it cannot be simplified. For example, an answer of √ 28 will be given in simplified form as 2√ 7. This option is useful for algebra 1 and 2 courses.

  12. 1.2: Square Root Property, Complete the Square, and The Quadratic

    Solve Quadratic Equations by Using the Square Root Property. A quadratic equation in standard form is \ (a x ^ { 2 } + b x + c = 0\) where \ (a, b\), and \ (c\) are real numbers and \ (a ≠ 0\). Quadratic equations can have two real solutions, one real solution, or no real solution—in which case there will be two complex solutions.

  13. Solving square-root equations: one solution

    Solving square-root equations: one solution Google Classroom About Transcript Sal solves the equation 3+√ (5x+6)=12. Created by Sal Khan and Monterey Institute for Technology and Education. Questions Tips & Thanks Want to join the conversation? Sort by: Top Voted Gillian Bush 12 years ago What does he mean by "Principle square root"?

  14. Solving Quadratic Equations by Square Root Method

    The solutions to this quadratic formula are [latex]x = 3 [/latex] and [latex]x = - \,3 [/latex]. Example 4: Solve the quadratic equation below using the Square Root Method. The two parentheses should not bother you at all. The fact remains that all variables come in the squared form, which is what we want. This problem is perfectly solvable ...

  15. Square Numbers and Square Roots Practice Questions

    Click here for Answers. Practice Questions. Previous: Irrational and Rational Numbers Practice Questions. Next: Expressing as a Percentage Practice Questions. The Corbettmaths Practice Questions on Square Numbers, Squaring Numbers and Square Roots.

  16. Square Root Questions

    1. Which of the following numbers is a perfect square? (a) 141 (b) 196 (c) 124 (d) 222 2. A perfect square number can never have the digit ….. at the units place. (a) 1 (b) 4 (c) 8 (d) 9

  17. Solving square-root equations: two solutions

    1) a=b (including the case a=0 and b=0) 2) a=-b (which is the same as -a=b) If there is a solution for -a=b, it is going to be extraneous to the original equation a=b. However, in this case we have 2 different values of w that satisfy our a=b case: -4 and -6.

  18. Square Roots and Cube Roots Questions [Solved]

    Square roots and cube roots questions with solutions are provided here for practice. Practising these questions will develop a deeper understanding of square and cube roots. The square root of a number means the given number is raised to the power ½, and it is denoted by the symbol '√' called the radical sign.

  19. Square Root Calculator

    Step 1: Enter the radical expression below for which you want to calculate the square root. The square root calculator finds the square root of the given radical expression. If a given number is a perfect square, you will get a final answer in exact form.

  20. 4.8: The Square Root Function

    Note the exact agreement with the graph of the square root function in Figure 1(c). The sequence of graphs in Figure 2 also help us identify the domain and range of the square root function. In Figure 2(a), the parabola opens outward indefinitely, both left and right. Consequently, the domain is \(D_{f} = (−\infty, \infty)\), or all real numbers.

  21. Square Roots and Cube Roots Questions

    Test Series. Understanding square roots and cube roots is an integral part of mathematics. Here, we provide you with a series of practice questions and solutions to help you gain a deeper knowledge of these concepts. A square root of a number is a value that, when multiplied by itself, gives the original number. It is denoted by the symbol '√'.

  22. Important Questions Class 8 Maths Chapter 6- Squares and Square Roots

    Class 8 Maths important questions for chapter 6-Squares and Square Roots are available here with solutions. Our subject experts have also provided the solutions for these problems based on CBSE latest syllabus and in reference with NCERT book.

  23. NCERT Solutions for Class 8 Maths Chapter 6 Squares and Square Roots

    Solution: The unit digit of square of a number having 'a' at its unit place ends with a×a. i. The unit digit of the square of a number having digit 1 as unit's place is 1. ∴ Unit digit of the square of number 81 is equal to 1. ii.