mathematical equations not solved

10 Math Equations That Have Never Been Solved

By Kathleen Cantor, 10 Sep 2020

Mathematics has played a major role in so many life-altering inventions and theories. But there are still some math equations that have managed to elude even the greatest minds, like Einstein and Hawkins. Other equations, however, are simply too large to compute. So for whatever reason, these puzzling problems have never been solved. But what are they?

Like the rest of us, you're probably expecting some next-level difficulty in these mathematical problems. Surprisingly, that is not the case. Some of these equations are even based on elementary school concepts and are easily understandable - just unsolvable.

1. The Riemann Hypothesis

Equation: σ (n) ≤ Hn +ln (Hn)eHn

• Where n is a positive integer
• Hn is the n-th harmonic number
• σ(n) is the sum of the positive integers divisible by n

For an instance, if n = 4 then σ(4)=1+2+4=7 and H4 = 1+1/2+1/3+1/4. Solve this equation to either prove or disprove the following inequality n≥1? Does it hold for all n≥1?

This problem is referred to as Lagarias’s Elementary Version of the Riemann Hypothesis and has a price of a million dollars offered by the  Clay Mathematics Foundation  for its solution.

2. The Collatz Conjecture

Equation: 3n+1

• where n is a positive integer n/2
• where n is a non-negative integer

Prove the answer end by cycling through 1,4,2,1,4,2,1,… if n is a positive integer. This is a repetitive process and you will repeat it with the new value of n you get. If your first n = 1 then your subsequent answers will be 1, 4, 2, 1, 4, 2, 1, 4… infinitely. And if n = 5 the answers will be 5,16,8,4,2,1 the rest will be another loop of the values 1, 4, and 2.

This equation was formed in 1937 by a man named Lothar Collatz which is why it is referred to as the Collatz Conjecture.

3. The Erdős-Strauss Conjecture

Equation: 4/n=1/a+1/b+1/c

• a, b and c are positive integers.

This equation aims to see if we can prove that for if n is greater than or equal to 2, then one can write 4*n as a sum of three positive unit fractions.

This equation was formed in 1948 by two men named Paul Erdős and Ernst Strauss which is why it is referred to as the Erdős-Strauss Conjecture.

4. Equation Four

Equation: Use 2(2∧127)-1 – 1 to prove or disprove if it’s a prime number or not?

Looks pretty straight forward, does it? Here is a little context on the problem.

Let’s take a prime number 2. Now, 22 – 1 = 3 which is also a prime number. 25 – 1 = 31 which is also a prime number and so is 27−1=127. 2127 −1=170141183460469231731687303715884105727 is also prime.

5. Goldbach's Conjecture

Equation: Prove that x + y = n

• where x and y are any two primes

This problem, as relatively simple as it sounds has never been solved. Solving this problem will earn you a free million dollars. This equation was first proposed by Goldbach hence the name Goldbach's Conjecture.

If you are still unsure then pick any even number like 6, it can also be expressed as 1 + 5, which is two primes. The same goes for 10 and 26.

6. Equation Six

Equation: Prove that (K)n = JK1N(q)JO1N(q)

• Where O = unknot (we are dealing with  knot theory )
• (K)n  =  Kashaev's invariant of K for any K or knot
• JK1N(q) of K is equal to N- colored Jones polynomial
• We also have the volume of conjecture as (EQ3)
• Here vol(K)  =  hyperbolic volume

This equation tries to portray the relationship between  quantum invariants  of knots and  the hyperbolic geometry  of  knot complements . Although this equation is in mathematics, you have to be a physics familiar to grasp the concept.

7. The Whitehead Conjecture

Equation: G = (S | R)

• when CW complex K (S | R) is aspherical
• if π2 (K (S | R)) = 0

What you are doing in this equation is prove the claim made by Mr.  Whitehead  in 1941 in  an algebraic topology  that every subcomplex of an  aspherical   CW complex  that is connected and in two dimensions is also spherical. This was named after the man, Whitehead conjecture.

8. Equation Eight

Equation: (EQ4)

• Where Γ = a  second countable   locally compact group
• And the * and r subscript = 0 or 1.

This equation is the definition of  morphism  and is referred to as an assembly map.  Check out the  reduced C*-algebra  for more insight into the concept surrounding this equation.

9. The Euler-Mascheroni Constant

Equation: y=limn→∞(∑m=1n1m−log(n))

Find out if y is rational or irrational in the equation above. To fully understand this problem you need to take another look at rational numbers and their concepts.  The character y is what is known as the Euler-Mascheroni constant and it has a value of 0.5772.

This equation has been calculated up to almost half of a trillion digits and yet no one has been able to tell if it is a rational number or not.

10. Equation Ten

Equation: π + e

Find the sum and determine if it is algebraic or transcendental. To understand this question you need to have an idea of  algebraic real numbers  and how they operate. The number pi or π originated in the 17th century and it is transcendental along with e. but what about their sum? So Far this has never been solved.

As you can see in the equations above, there are several seemingly simple mathematical equations and theories that have never been put to rest. Decades are passing while these problems remain unsolved. If you're looking for a brain teaser, finding the solutions to these problems will give you a run for your money.

See the 26 Comments below.

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10 Hard Math Problems That Continue to Stump Even the Brightest Minds

Maybe you’ll have better luck.

For now, you can take a crack at the hardest math problems known to man, woman, and machine. For more puzzles and brainteasers, check out Puzzmo . ✅ More from Popular Mechanics :

• To Create His Geometric Artwork, M.C. Escher Had to Learn Math the Hard Way
• Fourier Transforms: The Math That Made Color TV Possible
• The Game of Trees is a Mad Math Theory That Is Impossible to Prove

The Collatz Conjecture

In September 2019, news broke regarding progress on this 82-year-old question, thanks to prolific mathematician Terence Tao. And while the story of Tao’s breakthrough is promising, the problem isn’t fully solved yet.

A refresher on the Collatz Conjecture : It’s all about that function f(n), shown above, which takes even numbers and cuts them in half, while odd numbers get tripled and then added to 1. Take any natural number, apply f, then apply f again and again. You eventually land on 1, for every number we’ve ever checked. The Conjecture is that this is true for all natural numbers (positive integers from 1 through infinity).

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Tao’s recent work is a near-solution to the Collatz Conjecture in some subtle ways. But he most likely can’t adapt his methods to yield a complete solution to the problem, as Tao subsequently explained. So, we might be working on it for decades longer.

The Conjecture lives in the math discipline known as Dynamical Systems , or the study of situations that change over time in semi-predictable ways. It looks like a simple, innocuous question, but that’s what makes it special. Why is such a basic question so hard to answer? It serves as a benchmark for our understanding; once we solve it, then we can proceed onto much more complicated matters.

The study of dynamical systems could become more robust than anyone today could imagine. But we’ll need to solve the Collatz Conjecture for the subject to flourish.

Goldbach’s Conjecture

One of the greatest unsolved mysteries in math is also very easy to write. Goldbach’s Conjecture is, “Every even number (greater than two) is the sum of two primes.” You check this in your head for small numbers: 18 is 13+5, and 42 is 23+19. Computers have checked the Conjecture for numbers up to some magnitude. But we need proof for all natural numbers.

Goldbach’s Conjecture precipitated from letters in 1742 between German mathematician Christian Goldbach and legendary Swiss mathematician Leonhard Euler , considered one of the greatest in math history. As Euler put it, “I regard [it] as a completely certain theorem, although I cannot prove it.”

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Euler may have sensed what makes this problem counterintuitively hard to solve. When you look at larger numbers, they have more ways of being written as sums of primes, not less. Like how 3+5 is the only way to break 8 into two primes, but 42 can broken into 5+37, 11+31, 13+29, and 19+23. So it feels like Goldbach’s Conjecture is an understatement for very large numbers.

Still, a proof of the conjecture for all numbers eludes mathematicians to this day. It stands as one of the oldest open questions in all of math.

The Twin Prime Conjecture

Together with Goldbach’s, the Twin Prime Conjecture is the most famous in Number Theory—or the study of natural numbers and their properties, frequently involving prime numbers. Since you've known these numbers since grade school, stating the conjectures is easy.

When two primes have a difference of 2, they’re called twin primes. So 11 and 13 are twin primes, as are 599 and 601. Now, it's a Day 1 Number Theory fact that there are infinitely many prime numbers. So, are there infinitely many twin primes? The Twin Prime Conjecture says yes.

Let’s go a bit deeper. The first in a pair of twin primes is, with one exception, always 1 less than a multiple of 6. And so the second twin prime is always 1 more than a multiple of 6. You can understand why, if you’re ready to follow a bit of heady Number Theory.

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All primes after 2 are odd. Even numbers are always 0, 2, or 4 more than a multiple of 6, while odd numbers are always 1, 3, or 5 more than a multiple of 6. Well, one of those three possibilities for odd numbers causes an issue. If a number is 3 more than a multiple of 6, then it has a factor of 3. Having a factor of 3 means a number isn’t prime (with the sole exception of 3 itself). And that's why every third odd number can't be prime.

How’s your head after that paragraph? Now imagine the headaches of everyone who has tried to solve this problem in the last 170 years.

The good news is that we’ve made some promising progress in the last decade. Mathematicians have managed to tackle closer and closer versions of the Twin Prime Conjecture. This was their idea: Trouble proving there are infinitely many primes with a difference of 2? How about proving there are infinitely many primes with a difference of 70,000,000? That was cleverly proven in 2013 by Yitang Zhang at the University of New Hampshire.

For the last six years, mathematicians have been improving that number in Zhang’s proof, from millions down to hundreds. Taking it down all the way to 2 will be the solution to the Twin Prime Conjecture. The closest we’ve come —given some subtle technical assumptions—is 6. Time will tell if the last step from 6 to 2 is right around the corner, or if that last part will challenge mathematicians for decades longer.

The Riemann Hypothesis

Today’s mathematicians would probably agree that the Riemann Hypothesis is the most significant open problem in all of math. It’s one of the seven Millennium Prize Problems , with $1 million reward for its solution. It has implications deep into various branches of math, but it’s also simple enough that we can explain the basic idea right here.

There is a function, called the Riemann zeta function, written in the image above.

For each s, this function gives an infinite sum, which takes some basic calculus to approach for even the simplest values of s. For example, if s=2, then 𝜁(s) is the well-known series 1 + 1/4 + 1/9 + 1/16 + …, which strangely adds up to exactly 𝜋²/6. When s is a complex number—one that looks like a+b𝑖, using the imaginary number 𝑖—finding 𝜁(s) gets tricky.

So tricky, in fact, that it’s become the ultimate math question. Specifically, the Riemann Hypothesis is about when 𝜁(s)=0; the official statement is, “Every nontrivial zero of the Riemann zeta function has real part 1/2.” On the plane of complex numbers, this means the function has a certain behavior along a special vertical line. The hypothesis is that the behavior continues along that line infinitely.

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The Hypothesis and the zeta function come from German mathematician Bernhard Riemann, who described them in 1859. Riemann developed them while studying prime numbers and their distribution. Our understanding of prime numbers has flourished in the 160 years since, and Riemann would never have imagined the power of supercomputers. But lacking a solution to the Riemann Hypothesis is a major setback.

If the Riemann Hypothesis were solved tomorrow, it would unlock an avalanche of further progress. It would be huge news throughout the subjects of Number Theory and Analysis. Until then, the Riemann Hypothesis remains one of the largest dams to the river of math research.

The Birch and Swinnerton-Dyer Conjecture

The Birch and Swinnerton-Dyer Conjecture is another of the six unsolved Millennium Prize Problems, and it’s the only other one we can remotely describe in plain English. This Conjecture involves the math topic known as Elliptic Curves.

When we recently wrote about the toughest math problems that have been solved , we mentioned one of the greatest achievements in 20th-century math: the solution to Fermat’s Last Theorem. Sir Andrew Wiles solved it using Elliptic Curves. So, you could call this a very powerful new branch of math.

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In a nutshell, an elliptic curve is a special kind of function. They take the unthreatening-looking form y²=x³+ax+b. It turns out functions like this have certain properties that cast insight into math topics like Algebra and Number Theory.

British mathematicians Bryan Birch and Peter Swinnerton-Dyer developed their conjecture in the 1960s. Its exact statement is very technical, and has evolved over the years. One of the main stewards of this evolution has been none other than Wiles. To see its current status and complexity, check out this famous update by Wells in 2006.

The Kissing Number Problem

A broad category of problems in math are called the Sphere Packing Problems. They range from pure math to practical applications, generally putting math terminology to the idea of stacking many spheres in a given space, like fruit at the grocery store. Some questions in this study have full solutions, while some simple ones leave us stumped, like the Kissing Number Problem.

When a bunch of spheres are packed in some region, each sphere has a Kissing Number, which is the number of other spheres it’s touching; if you’re touching 6 neighboring spheres, then your kissing number is 6. Nothing tricky. A packed bunch of spheres will have an average kissing number, which helps mathematically describe the situation. But a basic question about the kissing number stands unanswered.

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First, a note on dimensions. Dimensions have a specific meaning in math: they’re independent coordinate axes. The x-axis and y-axis show the two dimensions of a coordinate plane. When a character in a sci-fi show says they’re going to a different dimension, that doesn’t make mathematical sense. You can’t go to the x-axis.

A 1-dimensional thing is a line, and 2-dimensional thing is a plane. For these low numbers, mathematicians have proven the maximum possible kissing number for spheres of that many dimensions. It’s 2 when you’re on a 1-D line—one sphere to your left and the other to your right. There’s proof of an exact number for 3 dimensions, although that took until the 1950s.

Beyond 3 dimensions, the Kissing Problem is mostly unsolved. Mathematicians have slowly whittled the possibilities to fairly narrow ranges for up to 24 dimensions, with a few exactly known, as you can see on this chart . For larger numbers, or a general form, the problem is wide open. There are several hurdles to a full solution, including computational limitations. So expect incremental progress on this problem for years to come.

The Unknotting Problem

The simplest version of the Unknotting Problem has been solved, so there’s already some success with this story. Solving the full version of the problem will be an even bigger triumph.

You probably haven’t heard of the math subject Knot Theory . It ’s taught in virtually no high schools, and few colleges. The idea is to try and apply formal math ideas, like proofs, to knots, like … well, what you tie your shoes with.

For example, you might know how to tie a “square knot” and a “granny knot.” They have the same steps except that one twist is reversed from the square knot to the granny knot. But can you prove that those knots are different? Well, knot theorists can.

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Knot theorists’ holy grail problem was an algorithm to identify if some tangled mess is truly knotted, or if it can be disentangled to nothing. The cool news is that this has been accomplished! Several computer algorithms for this have been written in the last 20 years, and some of them even animate the process .

But the Unknotting Problem remains computational. In technical terms, it’s known that the Unknotting Problem is in NP, while we don ’ t know if it’s in P. That roughly means that we know our algorithms are capable of unknotting knots of any complexity, but that as they get more complicated, it starts to take an impossibly long time. For now.

If someone comes up with an algorithm that can unknot any knot in what’s called polynomial time, that will put the Unknotting Problem fully to rest. On the flip side, someone could prove that isn’t possible, and that the Unknotting Problem’s computational intensity is unavoidably profound. Eventually, we’ll find out.

The Large Cardinal Project

If you’ve never heard of Large Cardinals , get ready to learn. In the late 19th century, a German mathematician named Georg Cantor figured out that infinity comes in different sizes. Some infinite sets truly have more elements than others in a deep mathematical way, and Cantor proved it.

There is the first infinite size, the smallest infinity , which gets denoted ℵ₀. That’s a Hebrew letter aleph; it reads as “aleph-zero.” It’s the size of the set of natural numbers, so that gets written |ℕ|=ℵ₀.

Next, some common sets are larger than size ℵ₀. The major example Cantor proved is that the set of real numbers is bigger, written |ℝ|>ℵ₀. But the reals aren’t that big; we’re just getting started on the infinite sizes.

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For the really big stuff, mathematicians keep discovering larger and larger sizes, or what we call Large Cardinals. It’s a process of pure math that goes like this: Someone says, “I thought of a definition for a cardinal, and I can prove this cardinal is bigger than all the known cardinals.” Then, if their proof is good, that’s the new largest known cardinal. Until someone else comes up with a larger one.

Throughout the 20th century, the frontier of known large cardinals was steadily pushed forward. There’s now even a beautiful wiki of known large cardinals , named in honor of Cantor. So, will this ever end? The answer is broadly yes, although it gets very complicated.

In some senses, the top of the large cardinal hierarchy is in sight. Some theorems have been proven, which impose a sort of ceiling on the possibilities for large cardinals. But many open questions remain, and new cardinals have been nailed down as recently as 2019. It’s very possible we will be discovering more for decades to come. Hopefully we’ll eventually have a comprehensive list of all large cardinals.

What’s the Deal with 𝜋+e?

Given everything we know about two of math’s most famous constants, 𝜋 and e , it’s a bit surprising how lost we are when they’re added together.

This mystery is all about algebraic real numbers . The definition: A real number is algebraic if it’s the root of some polynomial with integer coefficients. For example, x²-6 is a polynomial with integer coefficients, since 1 and -6 are integers. The roots of x²-6=0 are x=√6 and x=-√6, so that means √6 and -√6 are algebraic numbers.

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All rational numbers, and roots of rational numbers, are algebraic. So it might feel like “most” real numbers are algebraic. Turns out, it’s actually the opposite. The antonym to algebraic is transcendental, and it turns out almost all real numbers are transcendental—for certain mathematical meanings of “almost all.” So who’s algebraic , and who’s transcendental?

The real number 𝜋 goes back to ancient math, while the number e has been around since the 17th century. You’ve probably heard of both, and you’d think we know the answer to every basic question to be asked about them, right?

Well, we do know that both 𝜋 and e are transcendental. But somehow it’s unknown whether 𝜋+e is algebraic or transcendental. Similarly, we don’t know about 𝜋e, 𝜋/e, and other simple combinations of them. So there are incredibly basic questions about numbers we’ve known for millennia that still remain mysterious.

Is 𝛾 Rational?

Here’s another problem that’s very easy to write, but hard to solve. All you need to recall is the definition of rational numbers.

Rational numbers can be written in the form p/q, where p and q are integers. So, 42 and -11/3 are rational, while 𝜋 and √2 are not. It’s a very basic property, so you’d think we can easily tell when a number is rational or not, right?

Meet the Euler-Mascheroni constant 𝛾, which is a lowercase Greek gamma. It’s a real number, approximately 0.5772, with a closed form that’s not terribly ugly; it looks like the image above.

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The sleek way of putting words to those symbols is “gamma is the limit of the difference of the harmonic series and the natural log.” So, it’s a combination of two very well-understood mathematical objects. It has other neat closed forms, and appears in hundreds of formulas.

But somehow, we don’t even know if 𝛾 is rational. We’ve calculated it to half a trillion digits, yet nobody can prove if it’s rational or not. The popular prediction is that 𝛾 is irrational. Along with our previous example 𝜋+e, we have another question of a simple property for a well-known number, and we can’t even answer it.

Dave Linkletter is a Ph.D. candidate in Pure Mathematics at the University of Nevada, Las Vegas. His research is in Large Cardinal Set Theory. He also teaches undergrad classes, and enjoys breaking down popular math topics for wide audiences.

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Unsolved Problems

There are many unsolved problems in mathematics. Some prominent outstanding unsolved problems (as well as some which are not necessarily so well known) include

1. The Goldbach conjecture .

2. The Riemann hypothesis .

3. The conjecture that there exists a Hadamard matrix for every positive multiple of 4.

4. The twin prime conjecture (i.e., the conjecture that there are an infinite number of twin primes ).

5. Determination of whether NP-problems are actually P-problems .

6. The Collatz problem .

7. Proof that the 196-algorithm does not terminate when applied to the number 196.

8. Proof that 10 is a solitary number .

11. Finding an Euler brick whose space diagonal is also an integer.

12. Proving which numbers can be represented as a sum of three or four (positive or negative) cubic numbers .

14. Determining if the Euler-Mascheroni constant is irrational .

15. Deriving an analytic form for the square site percolation threshold .

16. Determining if any odd perfect numbers exist.

The Clay Mathematics Institute ( http://www.claymath.org/millennium/ ) of Cambridge, Massachusetts (CMI) has named seven "Millennium Prize Problems," selected by focusing on important classic questions in mathematics that have resisted solution over the years. A $7 million prize fund has been established for the solution to these problems, with $1 million allocated to each. The problems consist of the Riemann hypothesis , Poincaré conjecture , Hodge conjecture , Swinnerton-Dyer Conjecture , solution of the Navier-Stokes equations, formulation of Yang-Mills theory, and determination of whether NP-problems are actually P-problems .

In 1900, David Hilbert proposed a list of 23 outstanding problems in mathematics ( Hilbert's problems ), a number of which have now been solved, but some of which remain open. In 1912, Landau proposed four simply stated problems, now known as Landau's problems , which continue to defy attack even today. One hundred years after Hilbert, Smale (2000) proposed a list of 18 outstanding problems.

K. S. Brown, D. Eppstein, S. Finch, and C. Kimberling maintain webpages of unsolved problems in mathematics. Classic texts on unsolved problems in various areas of mathematics are Croft et al. (1991), in geometry , and Guy (2004), in number theory .

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Weisstein, Eric W. "Unsolved Problems." From MathWorld --A Wolfram Web Resource. https://mathworld.wolfram.com/UnsolvedProblems.html

Subject classifications

Solving Equations

What is an equation.

An equation says that two things are equal. It will have an equals sign "=" like this:

That equations says:

what is on the left (x − 2)  equals  what is on the right (4)

So an equation is like a statement " this equals that "

What is a Solution?

A Solution is a value we can put in place of a variable (such as x ) that makes the equation true .

Example: x − 2 = 4

When we put 6 in place of x we get:

which is true

So x = 6 is a solution.

How about other values for x ?

• For x=5 we get "5−2=4" which is not true , so x=5 is not a solution .
• For x=9 we get "9−2=4" which is not true , so x=9 is not a solution .

In this case x = 6 is the only solution.

You might like to practice solving some animated equations .

More Than One Solution

There can be more than one solution.

Example: (x−3)(x−2) = 0

When x is 3 we get:

(3−3)(3−2) = 0 × 1 = 0

And when x is 2 we get:

(2−3)(2−2) = (−1) × 0 = 0

which is also true

So the solutions are:

x = 3 , or x = 2

When we gather all solutions together it is called a Solution Set

The above solution set is: {2, 3}

Solutions Everywhere!

Some equations are true for all allowed values and are then called Identities

Example: sin(−θ) = −sin(θ) is one of the Trigonometric Identities

Let's try θ = 30°:

sin(−30°) = −0.5 and

−sin(30°) = −0.5

So it is true for θ = 30°

Let's try θ = 90°:

sin(−90°) = −1 and

−sin(90°) = −1

So it is also true for θ = 90°

Is it true for all values of θ ? Try some values for yourself!

How to Solve an Equation

There is no "one perfect way" to solve all equations.

A Useful Goal

But we often get success when our goal is to end up with:

x = something

In other words, we want to move everything except "x" (or whatever name the variable has) over to the right hand side.

Example: Solve 3x−6 = 9

Now we have x = something ,

and a short calculation reveals that x = 5

Like a Puzzle

In fact, solving an equation is just like solving a puzzle. And like puzzles, there are things we can (and cannot) do.

Here are some things we can do:

• Add or Subtract the same value from both sides
• Clear out any fractions by Multiplying every term by the bottom parts
• Divide every term by the same nonzero value
• Combine Like Terms
• Expanding (the opposite of factoring) may also help
• Recognizing a pattern, such as the difference of squares
• Sometimes we can apply a function to both sides (e.g. square both sides)

Example: Solve √(x/2) = 3

And the more "tricks" and techniques you learn the better you will get.

Special Equations

There are special ways of solving some types of equations. Learn how to ...

• solve Quadratic Equations
• solve Radical Equations
• solve Equations with Sine, Cosine and Tangent

Check Your Solutions

You should always check that your "solution" really is a solution.

How To Check

Take the solution(s) and put them in the original equation to see if they really work.

Example: solve for x:

2x x − 3 + 3 = 6 x − 3     (x≠3)

We have said x≠3 to avoid a division by zero.

Let's multiply through by (x − 3) :

2x + 3(x−3) = 6

Bring the 6 to the left:

2x + 3(x−3) − 6 = 0

Expand and solve:

2x + 3x − 9 − 6 = 0

5x − 15 = 0

5(x − 3) = 0

Which can be solved by having x=3

Let us check x=3 using the original question:

2 × 3 3 − 3 + 3  =   6 3 − 3

Hang On: 3 − 3 = 0 That means dividing by Zero!

And anyway, we said at the top that x≠3 , so ...

x = 3 does not actually work, and so:

There is No Solution!

That was interesting ... we thought we had found a solution, but when we looked back at the question we found it wasn't allowed!

This gives us a moral lesson:

"Solving" only gives us possible solutions, they need to be checked!

• Note down where an expression is not defined (due to a division by zero, the square root of a negative number, or some other reason)
• Show all the steps , so it can be checked later (by you or someone else)

An equation is a statement of the equality of two expressions . An equation is formed when an equals sign is placed between two expressions. An algebraic equation is an equation that includes a variable.

Equations differ from expressions in that expressions can only be simplified or evaluated for a given value, not solved. Equations on the other hand, can be solved. Solving an equation involves finding the values for which the equation is true. Solving equations by learning how to manipulate the quantities or expressions in an equation is a large part of elementary algebra, which is used on some level in almost all areas of mathematics.

Below is an example of an equation.

x + 3 = 2x + 4

The solution of the equation is x = -1. We can confirm this by plugging -1 in for x:

-1 + 3 = 2(-1) + 4

Since plugging -1 in for x makes a true statement, we know that -1 is a solution to the equation. It is possible for an equation to have no solution.

x + 2 = x + 1

The above equation has no solution because there is no value for x such that the equation can be true. For the above equation to be true, 2 = 1 would have to be true, which is clearly not the case. Determining that an equation has no solution also constitutes solving the equation.

Equations can also have more than one solution. For example:

(x + 3)(x - 5) = 0

This equation has two solutions: x = -3 and x = 5. We can determine this by first dividing each side of the equation by one of the terms, which leaves the other term. Using basic arithmetic, we can then solve for the term, after which we repeat the process for the other term:

An equation that is true for all values of a variable is known as an identity. There are numerous identities in mathematics. Trigonometry is one branch of mathematics where identities are commonly used.

The above are some of the more basic aspects of equations. There are many different types or classifications of equations. Some examples include quadratic equations, polynomial equations, linear equations, parametric equations, differential equations, and many more. In the context of this section of the site, we will mostly be working with various types of algebraic equations where the main purpose is to solve the equation.

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The equations section of QuickMath allows you to solve and plot virtually any equation or system of equations. In most cases, you can find exact solutions to your equations. Even when this is not possible, QuickMath may be able to give you approximate solutions to almost any level of accuracy you require. It also contains a number of special commands for dealing with quadratic equations.

The Solve command can be uses to solve either a single equation for a single unknown from the basic solve page or to simultaneously solve a system of many equations in many unknowns from the advanced solve page . The advanced command allows you to specify whether you want approximate numerical answers as well as exact ones, and how many digits of accuracy (up to 16) you require. It also allows you to eliminate certain variables from the equations.

Go to the Solve page

The Plot command, from the Graphs section, will plot any function of two variables. In order to plot a single function of x, go to the basic equation plotting page , where you can enter the equation and specify the upper and lower limits on x that you want the graph to be plotted for. The advanced plotting page allows you to plot up to 6 equations on the one graph, each with their own color. It also gives you control over such things as whether or not to show the axes, where the axes should be located, what the aspect ratio of the plot should be and what the range of the dependent variable should be. All equations can be given in the explicit y = f(x) form or the implicit g(x,y) = c form.

Go to the Equation Plotting page

The Quadratics page contains 13 separate commands for dealing with the most common questions concerning quadratics. It allows you to : factor a quadratic function (by two different methods); solve a quadratic equation by factoring the quadratic, using the quadratic formula or by completing the square; rewrite a quadratic function in a different form by completing the square; calculate the concavity, x-intercepts, y-intercept, axis of symmetry and vertex of a parabola; plot a parabola; calculate the discriminant of a quadratic equation and use the discriminant to find the number of roots of a quadratic equation. Each command generates a complete and detailed custom-made explanation of all the steps needed to solve the problem.

Go to the Quadratics page

Introduction to Equations

By an equation we mean a mathematical sentence that states that two algebraic expressions are equal. For example, a (b + c) =ab + ac, ab = ba, and x 2 -1 = (x-1)(x+1) are all equations that we have been using. We recall that we defined a variable as a letter that may be replaced by numbers out of a given set, during a given discussion. This specified set of numbers is sometimes called the replacement set. In this chapter we will deal with equations involving variables where the replacement set, unless otherwise specified, is the set of all real numbers for which all the expressions in the equation are defined.

If an equation is true after the variable has been replaced by a specific number, then the number is called a solution of the equation and is said to satisfy it. Obviously, every solution is a member of the replacement set. The real number 3 is a solution of the equation 2x-1 = x+2, since 2*3-1=3+2. while 1 is a solution of the equation (x-1)(x+2) = 0. The set of all solutions of an equation is called the solution set of the equation.

In the first equation above {3} is the solution set, while in the second example {-2,1} is the solution set. We can verify by substitution that each of these numbers is a solution of its respective equation, and we will see later that these are the only solutions.

A conditional equation is an equation that is satisfied by some numbers from its replacement set and not satisfied by others. An identity is an equation that is satisfied by all numbers from its replacement set.

Example 1 Consider the equation 2x-1 = x+2

The replacement set here is the set of all real numbers. The equation is conditional since, for example, 1 is a member of the replacement set but not of the solution set.

Example 2 Consider the equation (x-1)(x+1) =x 2 -1 The replacement set is the set of all real numbers. From our laws of real numbers if a is any real number, then (a-1)(a+1) = a 2 -1 Therefore, every member of the replacement set is also a member of the solution set. Consequently this equation is an identity.  

The replacement set for this equation is the set of real numbers except 0, since 1/x and (1- x)/x are not defined for x = 0. If a is any real number in the replacement set, then

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4.2: Mathematical Expressions (NOT Equations)

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• Page ID 31001

• David Arnold
• College of the Redwoods

Recall the definition of a variable presented in Section 1.6.

Definition: Variable

A variable is a symbol (usually a letter) that stands for a value that may vary.

Let’s add the definition of a mathematical expression .

Definition: Mathematical Expression

When we combine numbers and variables in a valid way, using operations such as addition, subtraction, multiplication, division, exponentiation, and other operations and functions as yet unlearned, the resulting combination of mathematical symbols is called a mathematical expression .

2 a , x + 5, and y 2 ,

being formed by a combination of numbers, variables, and mathematical operators, are valid mathematical expressions. A mathematical expression must be well-formed . For example,

2 + ÷5 x

is not a valid expression because there is no term following the plus sign (it is not valid to write +÷ with nothing between these operators). Similarly,

is not well-formed because parentheses are not balanced.

Translating Words into Mathematical Expressions

In this section we turn our attention to translating word phrases into mathematical expressions. We begin with phrases that translate into sums . There is a wide variety of word phrases that translate into sums. Some common examples are given in Table \(\PageIndex{1a}\), though the list is far from complete. In like manner, a number of phrases that translate into differences are shown in Table \(\PageIndex{1b}\).

Let’s look at some examples, some of which translate into expressions involving sums, and some of which translate into expressions involving differences.

Translate the following phrases into mathematical expressions:

• "12 larger than x, "
• "11 less than y ," and
• " r decreased by 9."

Here are the translations.

• “12 larger than x” becomes x + 12.
• “11 less than y” becomes y − 11.
• “r decreased by 9” becomes r − 9.
• "13 more than x " and
• "12 fewer than y ".

(a) x + 13 and

(b) y − 12

Let W represent the width of the rectangle. The length of a rectangle is 4 feet longer than its width. Express the length of the rectangle in terms of its width W .

We know that the width of the rectangle is W . Because the length of the rectangle is 4 feet longer that the width, we must add 4 to the width to find the length.

\[ \begin{array}{c c c c c} \colorbox{cyan}{Length} & \text{is} & \colorbox{cyan}{4} & \text{more than} & \colorbox{cyan}{the width} \\ \text{Length} & = & 4 & + & W \end{array}\nonumber \]

Thus, the length of the rectangle, in terms of its width W , is 4 + W .

The width of a rectangle is 5 inches shorter than its length L . Express the width of the rectangle in terms of its length L .

L − 5

A string measures 15 inches is cut into two pieces. Let x represent the length of one of the resulting pieces. Express the length of the second piece in terms of the length x of the first piece.

The string has original length 15 inches. It is cut into two pieces and the first piece has length x . To find the length of the second piece, we must subtract the length of the first piece from the total length.

\[ \begin{array}{c c c c c} \colorbox{cyan}{Length of the second piece} & \text{is} & \colorbox{cyan}{Total length} & \text{minus} & \colorbox{cyan}{the length of the first piece} \\ \text{Length of the second piece} & = & 15 & - & x \end{array}\nonumber \]

Thus, the length of the second piece, in terms of the length x of the first piece, Answer: 12 + x is 15 − x .

A string is cut into two pieces, the first of which measures 12 inches. Express the total length of the string as a function of x , where x represents the length of the second piece of string.

There is also a wide variety of phrases that translate into products. Some examples are shown in Table 3.2(a), though again the list is far from complete. In like manner, a number of phrases translate into quotients, as shown in Table 3.2(b).

Let’s look at some examples, some of which translate into expressions involving products, and some of which translate into expressions involving quotients.

Translate the following phrases into mathematical expressions: (a) “11 times x ,” (b) “quotient of y and 4,” and (c) “twice a .”

Here are the translations. a) “11 times x ” becomes 11 x . b) “quotient of y and 4” becomes y /4, or equivalently, \(\frac{y}{4}\). c) “twice a ” becomes 2 a .

Translate into mathematical symbols: (a) “the product of 5 and x ” and (b) “12 divided by y .”

(a) 5 x and (b) 12/ y .

A plumber has a pipe of unknown length x . He cuts it into 4 equal pieces. Find the length of each piece in terms of the unknown length x .

The total length is unknown and equal to x . The plumber divides it into 4 equal pieces. To find the length of each pieces, we must divide the total length by 4.

\[ \begin{array}{c c c c c} \colorbox{cyan}{Length of each piece} & \text{is} & \colorbox{cyan}{Total length} & \text{divided by} & \colorbox{cyan}{4} \\ \text{Length of each piece} & = & x & \div & 4 \end{array}\nonumber \]

Thus, the length of each piece, in terms of the unknown length x , is x /4, or equivalently, \(\frac{x}{4}\).

A carpenter cuts a board of unknown length L into three equal pieces. Express the length of each piece in terms of L .

Mary invests A dollars in a savings account paying 2% interest per year. She invests five times this amount in a certificate of deposit paying 5% per year. How much does she invest in the certificate of deposit, in terms of the amount A in the savings account?

The amount in the savings account is A dollars. She invests five times this amount in a certificate of deposit.

\[ \begin{array}{c c c c c} \colorbox{cyan}{Amount in CD} & \text{is} & \colorbox{cyan}{5} & \text{times} & \colorbox{cyan}{Amount in savings} \\ \text{Amount in CD} & = & 5 & \cdot & A \end{array}\nonumber \]

Thus, the amount invested in the certificate of deposit, in terms of the amount A in the savings account, is 5 A .

David invests K dollars in a savings account paying 3% per year. He invests half this amount in a mutual fund paying 4% per year. Express the amount invested in the mutual fund in terms of K , the amount invested in the savings account.

\(\frac{1}{2}K\)

Combinations

Some phrases require combinations of the mathematical operations employed in previous examples.

Let the first number equal x . The second number is 3 more than twice the first number. Express the second number in terms of the first number x .

The first number is x . The second number is 3 more than twice the first number.

\[ \begin{aligned} \colorbox{cyan}{Second number} & \text{is} & \colorbox{cyan}{3} & \text{more than} & \colorbox{cyan}{twice the first number} \\ \text{Second number} & = & 3 & + & 2x \end{aligned}\nonumber \]

Therefore, the second number, in terms of the first number x , is 3 + 2 x .

A second number is 4 less than 3 times a first number. Express the second number in terms of the first number y .

3 y − 4

The length of a rectangle is L . The width is 15 feet less than 3 times the length. What is the width of the rectangle in terms of the length L ?

The length of the rectangle is L . The width is 15 feet less than 3 times the length.

\[ \begin{aligned} \colorbox{cyan}{Width} & \text{is} & \colorbox{cyan}{3 times the length} & \text{less} & \colorbox{cyan}{15} \\ \text{Width} & = & 3L & - & 15 \end{aligned}\nonumber \]

Therefore, the width, in terms of the length L , is 3 L − 15.

The width of a rectangle is W . The length is 7 inches longer than twice the width. Express the length of the rectangle in terms of its length L .

In Exercises 1-20, translate the phrase into a mathematical expression involving the given variable.

1. “8 times the width n ”

2. “2 times the length z ”

3. “6 times the sum of the number n and 3”

4. “10 times the sum of the number n and 8”

5. “the demand b quadrupled”

6. “the supply y quadrupled”

7. “the speed y decreased by 33”

8. “the speed u decreased by 30”

9. “10 times the width n ”

10. “10 times the length z ”

11. “9 times the sum of the number z and 2”

12. “14 times the sum of the number n and 10”

13. “the supply y doubled”

14. “the demand n quadrupled”

15. “13 more than 15 times the number p ”

16. “14 less than 5 times the number y ”

17. “4 less than 11 times the number x ”

18. “13 less than 5 times the number p ”

19. “the speed u decreased by 10”

20. “the speed w increased by 32”

21. Representing Numbers. Suppose n represents a whole number.

i) What does n + 1 represent?

ii) What does n + 2 represent?

iii) What does n − 1 represent?

22. Suppose 2n represents an even whole number. How could we represent the next even number after 2n?

23. Suppose 2n + 1 represents an odd whole number. How could we represent the next odd number after 2n + 1?

24. There are b bags of mulch produced each month. How many bags of mulch are produced each year?

25. Steve sells twice as many products as Mike. Choose a variable and write an expression for each man’s sales.

26. Find a mathematical expression to represent the values.

i) How many quarters are in d dollars?

ii) How many minutes are in h hours?

iii) How many hours are in d days?

iv) How many days are in y years?

v) How many months are in y years?

vi) How many inches are in f feet?

vii) How many feet are in y yards?

3. 6(n + 3)

7. y − 33

11. 9(z + 2)

15. 15p + 13 17.

11x − 4

19. u − 10

i) n+1 represents the next whole number after n.

ii) n+2 represents the next whole number after n + 1, or, two whole numbers after n.

iii) n − 1 represents the whole number before n.

25. Let Mike sell p products. Then Steve sells 2p products.

When it comes to math, maybe the kids aren’t the problem

The way we are teaching, testing, and requiring mathematics is damaging our society with potentially fatal results. That’s a strong statement, I know, but to get a sense of why, take a moment to think of something that you would find hard or maybe impossible to do — executing 30 pullups, playing a violin concerto, walking a tightrope over an abyss, feeding a snake, solving a complex polynomial equation. 

Now suppose our society decrees that to graduate from high school, you must be able to do that thing or be branded a failure and not earn a diploma. How are you feeling about yourself and your prospects?

Imagine trying your hardest to learn the skills, but the textbooks, online resources and sometimes even teachers confuse you, and on top of that, some exams are so badly designed they are virtually impossible to pass. 

The 2022 results from the National Assessment of Educational Progress showed only 36% of fourth graders and 26% of eighth graders were considered proficient in mathematics. The national average for high school math proficiency is just 38%.

Why are so many young people not achieving math proficiency? Consider how we define “proficiency.” 

For example, the Common Core math standards state that, among other things, high school students should “Recognize that sequences are functions, sometimes defined recursively, whose domain is a subset of the integers. For example, the Fibonacci sequence is defined recursively by f (0) = f (1) = 1, f (n+1) = f (n) + f (n-1) for n ≥ 1.” And that’s just one example of many.

Suppose your child came to you with that as homework each night, possibly in tears. Now imagine that your child has special needs and deals with dyscalculia , ADD or other challenges.  

One of my sons struggles with ADD and is trying to complete an online financial algebra course offered by his school district to fulfill his mandatory “math pathway” for graduation. To help him succeed, a certified math and science tutor and I have worked with him every day through some very poorly designed online lessons provided by his district. 

We also helped him try to solve problems during a unit exam, but together, we achieved only a score of 32% on the exam. A certified math teacher, and a former statistics instructor with a Ph.D., failed a high school financial algebra exam. Why? Because the lessons were deeply flawed, the exam items were ambiguously worded, there were too many problems for the allotted time and the rules prohibited the use of resources like a financial calculator that one might enlist in real life. 

All that is not an excuse, and this column is not some reflexive “anti-testing” rant. It is, unfortunately, the truth. That leaves us asking, “If we failed the exam, how will most students, or their parents, possibly succeed?”

If you make children and their parents feel like failures, it is predictable that some are eventually going to give up on school and on themselves. Once that happens, they are far more likely to turn to behaviors that are harmful, self-destructive, and, tragically, sometimes fatal. And when significant portions of our population share those feelings, it inevitably harms our society as a whole. It is nothing less than educational and social malpractice to create unrealistic and unnecessary demands for children and then fail to provide them with the best resources to help them succeed.   

A good start would be to insist that policymakers, superintendents, school boards and school principals throughout our state attempt to learn each subject in math through the same texts or online resources students are required to use. I don’t mean just skimming a text or taking the word of someone else, I mean really digging in and trying to learn the material and take the tests firsthand. I doubt most people would accept the challenge and I am certain even fewer would pass the exams. 

That is not because the administrators and teachers aren’t smart, good people — it is because the materials and exams are so poorly designed, and the requirements are so profoundly detached from daily life. But if that is so, which the evidence suggests it is, why are we inflicting such things on our children and with what impacts?

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Students ask their maths teacher to solve an equation. The solution is heartwarming

A video showing a group of students at a school in nepal asking their maths teacher to solve a special equation went viral on instagram. it was shared on the platform by the account, 'class12diaries'..

Listen to Story

• Nepal school students ask their maths teacher to solve a unique equation
• Teacher solves the equation and the solution comes out as 'We Love You'
• Video of the scene goes viral on Instagram with around 10 million views

A touching moment captured at Motherland Secondary School in Nepal's Pokhara has caught the attention of social media users. A video showing a group of students asking their maths teacher to solve a special equation went viral after it was shared on Instagram by the account, 'Class12Diaries'.

The video features a student writing an algebraic expression on the board and requesting their teacher to solve it. The equation, crafted by the students, leads to an unexpected yet touching result that leaves both the students in class and the viewers in awe.

As the teacher works through the problem, he is initially puzzled, but soon discovers that the solution comes out as "We Love You." This sweet surprise was greeted with cheers from the students, visibly moved by their creation.

"We love (heart emoji) u," reads the caption of the post.

View this post on Instagram A post shared by 12B-io (@class12diaries)

The video has gone viral with over 10 million views.

According to the bio of the Instagram account, the page is managed by the students of Class 12B of Motherland Secondary School.

The video obtained positive reactions where people praised the students for their creativity. People also expressed their admiration for the touching moment between the teacher and his students.

"He won, he finally won with his subject," a person said. "OMG, I’m not crying, You’re crying," another commented.

The bond between a teacher and students weaves threads of inspiration and growth, thus creating a resilient fabric of shared knowledge and trust.

In another incident, a group of students paid special tribute to their teacher's son who died a year ago. A video of their heartfelt gesture left several social media users teary-eyed as they decided to honour the memory of her late son, Winston, in a profoundly moving way. Published By: Ashmita Saha Published On: Apr 6, 2024 READ | Viral video: Chhattisgarh students chase drunk teacher away, throw slippers at him

Google's AI search tool doesn't seem to have mastered math yet

• Google's AI search tool got simple mathematics wrong, a review by The Washington Post said.
• The experimental tool, Search Generative Experience, was introduced in May.
• Google is considering charging for some AI-powered features, The Financial Times reported.

Google's AI search tool appears to struggle when it comes to simple mathematics.

The tool was put to the test by Washington Post reporter Geoffrey A. Fowler, who found that despite extensive public testing, it stumbled over a straightforward question.

Google previously said it was "supercharging" and "improving" its search experience with a generative AI-infused version called Search Generative Experience (SGE).

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The AI question-answering tool is still an experiment. But as Fowler notes in a review published this week, it might have made it "dumber."

In the blog Google posted last May announcing the rollout, it said, "With new generative AI capabilities in Search, we're now taking more of the work out of searching, so you'll be able to understand a topic faster, uncover new viewpoints and insights, and get things done more easily."

Yet in Fowler's review, SGE fed back a response that was riddled with errors. The tech columnist searched for Meta CEO Mark Zuckerberg's net worth, and it replied, "$46.24 per hour, or $96,169 per year. This is equivalent to $8,014 per month, $1,849 per week, and $230.6 million per day."

The figures do not add up, which might be concerning given that Google is weighing up whether it should charge users for certain AI-powered features, The Financial Times reported.

Paying subscribers will rightly expect reliable answers, especially those focused on some of the world's most well-known business figures.

The company told the FT, "With our generative AI experiments in Search, we've already served billions of queries, and we're seeing positive Search query growth in all of our major markets. We're continuing to rapidly improve the product to serve new user needs."

Google already gives subscribers of its One AI Premium plan access to its AI model Gemini in features including Gmail, Docs, Slides, Sheets, and Meet.

But the company was forced to pull part of the Gemini AI model earlier this year after a series of blunders. Users complained the image-generating feature was creating historically inaccurate images of people of color.

Google didn't immediately respond to Business Insider's request for comment, made outside normal working hours.

On February 28, Axel Springer, Business Insider's parent company, joined 31 other media groups and filed a $2.3 billion suit against Google in Dutch court, alleging losses suffered due to the company's advertising practices.

Watch: Accenture CMO Jill Kramer talks about how generative AI will enhance, not diminish, the power of marketing: video

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How Tesla Planted the Seeds for Its Own Potential Downfall

Elon musk’s factory in china saved his company and made him ultrarich. now, it may backfire..

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When Elon Musk set up Tesla’s factory in China, he made a bet that brought him cheap parts and capable workers — a bet that made him ultrarich and saved his company.

Mara Hvistendahl, an investigative reporter for The Times, explains why, now, that lifeline may have given China the tools to beat Tesla at its own game.

On today’s episode

Mara Hvistendahl , an investigative reporter for The New York Times.

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A pivot to China saved Elon Musk. It also bound him to Beijing .

Mr. Musk helped create the Chinese electric vehicle industry. But he is now facing challenges there as well as scrutiny in the West over his reliance on China.

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Katrin Bennhold is the Berlin bureau chief. A former Nieman fellow at Harvard University, she previously reported from London and Paris, covering a range of topics from the rise of populism to gender. More about Katrin Bennhold

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