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## Algebra (all content)

### Course: Algebra (all content) > Unit 12

Lesson 3: Solving cube-root equations# Solving cube-root equations

Sal solves the equation -∛y=4∛y+5. Created by Sal Khan and Monterey Institute for Technology and Education.

## Want to join the conversation?

- are there any practice problems?(18 votes)
- Yes there are practice questions for this:

http://www.khanacademy.org/math/algebra/exponent-equations/radical_equations/e/radical_equations(7 votes)

- What is the difference between (-1^3) and (-1)^3?(4 votes)
- Although there is no difference in the result, there is a difference in the meaning of the two expressions.

One of them (-1^3) says that we have to

take the negative of`1 cubed`

So, we cube 1:

1 ∙ 1 ∙ 1 = 1

The negative of 1 is`-1`

The other one (-1)^3?

demands that we find the cube of`negative 1`

Taking the cube of -1:

(-1) ∙ (-1) ∙ (-1) =`-1`

So the answer is the same, but the process is different, or maybe I should say the pathway to the answer.(23 votes)

- Okay...for the last few days I've been obsessing over high level equations.. So let's get weedy here... First...I'm just trying to get a grasp mentally here. Why is it at a very basica school grade level reason, is it that higher degree equations (6+ it seems) are just literally unsolvable?

Also... these can't be solved for one varible like X....But what about euqations with two variables? Would that make the unsolvable, literally even harder?

Which would be more complex to solve, say a 4 degree polynomial, or a 3 degree but solving for two variables?

It sort of boggles my mind that in this age of super computers, computations and what not...we can't solve these equations?

Lastly...okay so we can't solve high degree equations, but is graphing them any easier?(4 votes)- First, understand that when we say polynomials of degree 5 and up are unsolvable, we mean 'not solvable by radicals'. That is, there is no formula involving only addition, subtraction, multiplication, division, integer roots, and integer exponents, into which we can plug the coefficients of a 5th-or-higher-degree polynomial, and which spits out all of the roots every time.

This doesn't mean we cannot find the roots of high-degree polynomials; we can. We just need different tools besides arithmetic and radicals. It's not a problem of computation, it's the fact that a certain type of mathematical object doesn't exist.

Why the general quintic is unsolvable isn't easy to explain at a low level. The course that covered it in my school (Galois theory) was third in a chain of prerequisites, starting with an intro to proofs course. But roughly, we study objects that concern permutations of the roots of a polynomial. A quadratic has only 2 roots, and only 2!=2 permutations. A cubic has 3 roots, so 3!=6 permutations. For the cubic, we manage to exploit some symmetries of the problem to reduce it to a quadratic equation. The quartic has 4 roots, and 4!=24 permutations, but we still manage to reduce it to a cubic equation by exploiting more symmetries. Then a quintic has 5 roots and 5!=120 permutations of the roots, and there just isn't enough symmetry to exploit to whittle the 120 permuations down to 24.

The thing to understand with 2-variable polynomials is that they are graphed as surfaces, not as curves. Our input space is the xy plane, and the polynomial is a wavy surface floating above/below it. The height of the surface above a point (x, y) is the value of the polynomial for that input (x, y), and the solution is the part of the surface that touches the xy plane. So generally, the solution of a 2-variable polynomial is a curve, not a small set of points. It's a whole different problem.

Lastly, the degree of a polynomial has little bearing on how hard it is to graph. A degree-10 polynomial is more tedious, but not more difficult to graph that a degree-5. You'll learn some techniques for drawing more accurate graphs in calculus.(8 votes)

- Has Sal ever checked the answer to a question in a video and found out he did it wrong?(5 votes)
- Wesim,

I think I have seen a couple KA videos where Sal checked his work part way through a video, realized he made an error and went back and fixed the error before continuing. They were small mistakes that were easy to go back and correct.

I suspect when he finds he made a big mistake, he just re-makes the video. Otherwise, someone watching only part of the video before the error might end up being confused.(6 votes)

- Why for the Radical Equations 1 did he multiply the 3 and in this he is dividing by 5 (1:22) ?(3 votes)
- Will solving a cube root equation get you a extraneous solution?(2 votes)
- No, cube roots don't give you two solutions by solving them. It is only when you raise it to an even number that it can give you an extraneous solution, because it it can not be reversed. You didn't get rid of any information, so you don't need to check for extraneous solutions unless your teacher makes you (for a bizzare reason), or it is to the power of an even number.(4 votes)

- At0:39, could he have added the cube root of y to each side instead of subtracting 4 times the cube root of y from each side?(2 votes)
- Yes, he could have. But then he would have another step to isolate the y. He would have to add the opposite of 5 to both sides (same as subtracting 5) to get the numeric term on the other side of the equation in preparation for the next step. Finally he would divide by positive 5 in this case in order to divide by the coefficient of the radical. Then he would be ready to cube both sides to reveal his answer.

-(y)^(1/3) = 4(y^(1/3)) + 5

+(y)^(1/3) + (y)^(1/3)

0 = 5(y^(1/3)) + 5

-5 -5

-5 = 5(y^(1/3))

-5/5 = 5(y^(1/3))/5

-1 = (y^(1/3))

(-1)³ = (y^(1/3))³

-1 = y

ta da!(2 votes)

- what if the numbers or variables inside the radical aren't the same?(2 votes)
- You wouldn't be able to solve the equation for a numerical value, but you could express one of the variables in terms of the other one. The only way to solve for both variables (that I know of) would be if you had a system of two equations or more.(2 votes)

- In this case, would we still need to check for extraneous solutions? Because there are no even powers or roots in this problem...(2 votes)
- No need to check for extraneous solutions, and you've identified exactly the reason why. We aren't raising both sides to an
**even**number, so it's reversible -- we can cube root both sides again and come back to the original function. Another way of saying it -- you haven't erased any information (e.g. negative signs) by raising both sides to the third power.(1 vote)

- wait how did you get -5 , isnt it supposed to be -4 instead(0 votes)
- The original equation says -1(y^1/3). When you subtract 4(y^1/3), you get (-1-4)(y^1/3) = -5(y^1/3).(8 votes)

## Video transcript

We're asked to solve for y. So we're told that the negative
of the cube root of y is equal to 4 times the
cube root of y plus 5. So in all of these it's helpful
to just be able to isolate the cube root, isolate
the radical in the equation, and then solve from there. So let's see if we can
isolate the radical. So the simplest thing to do, if
we want all of the radical onto the left-hand side
equation, we can subtract 4 times the cube root of y from
both sides of this equation. So let's subtract 4. We want to subtract 4 times the
cube root of y from both sides of this equation. And so your left-hand side, you
already have negative 1 times the cube root of y, and
you're going to subtract 4 more of the cube root of y. So you're going to have
negative 5 times the cube root of y. That's your left-hand side. Now the right-hand side-- these
two guys-- cancel out. That was the whole point behind
subtracting this value. So that cancels out and you're
just left with a 5 there. You're just left with this
5 right over there. Now, we've almost isolated
this cube root of y. We just have to divide both
sides of the equation by negative 5. So you just divide both
sides of this equation by negative 5. And these cancel out. That was the whole point. And we are left with the cube
root of y is equal to-- 5 divided by negative
5 is negative 1. Now, the cube root of y is
equal to negative 1. Well the easiest way to solve
this is, let's take both sides of this equation to
the third power. This statement right here is
the exact same statement as saying y to the 1/3 is
equal to negative 1. These are just two
different ways of writing the same thing. This is equivalent to taking
the 1/3 power. So if we take both sides of
this equation to the third power, that's like taking both
sides of this equation to the third power. And you can see here, y to the
1/3 to the third-- y to the 1/3 and then to the third,
that's like saying y to the 1/3 times 3 power. Or y to the first power. That's the whole point of it. If you take the cube root of y
to the third power, that's just going to be y. So the left-hand
side becomes y. And then the right-hand side,
what's negative 1 to the third power? Negative 1 times negative
1 is 1. Times negative 1 again
is negative 1. So we get y is equal to negative
1 as our solution. Now let's make sure that
it actually works. Let's go back to our
original equation. And I'll put negative
1 in for our y's. We had the negative of the
cube root of-- this time, negative 1-- has to be equal
to 4 times the cube root of negative 1 plus 5. Let's verify that this
is the case. The cube root of negative
1 is negative 1. Negative 1 to the third
power is negative 1. So this is equal to the negative
of negative 1 has to be equal to 4 times-- the cube
root of negative 1 is negative 1 plus 5. The negative of negative
1 is just positive 1. So 1 needs to be equal to--
4 times negative 1, negative 4, plus 5. This is true. Negative 4 plus 5 is 1. So this works out. This is our solution.