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### Course: Algebra (all content) > Unit 11

Lesson 7: Simplifying radicals (higher-index roots)# Simplifying higher-index roots

Multiple examples of simplifications of higher-index radicals. For example, simplifying ⁵√96 as 2⁵√3. Created by Sal Khan and CK-12 Foundation.

## Want to join the conversation?

- Can someone help me with something on the exercise

I understand it but I don't understand how to fill in the blank spots. The problem is this

They had the square root of 100 and it said siplify. 10 right but there are 2 blank spots what do I do?

my question is answered! thanks! feel free to vote up if you think it was a good question!(90 votes)- its plus and minus 10

+10 and -10

they both give you 100(57 votes)

- What is the principal cube root of -27?

I would expect it to be -3, but have been told it is actually 1.5+2.59808i. I can see that that is a solution to x^(1/3) = -27, but why is it considered 'principal root' rather than -3?(24 votes)- It is the principal root, so it must be positive. -3 is clearly not positive.(2 votes)

- Is there a faster way to do prime factorization(7 votes)
- To do "Prime Factorization", by definition, you are factoring all the way down as far as you can go, so NO, there is no faster way. However, sometimes you can recognize that a number doesn't have to be factored all the way down to its prime factors to be simplified. For example, if you were asked to simplify the square root of 96 (instead of the 5th root as in the problem above), you might recognize that 96 = 16 X 6 and that 16 is itself a perfect square (4 X 4). So, you wouldn't have to factor 96 all the way down to its prime factors of 2X2X2X2X2X3. You could take 4 (the square root of 16) out from underneath the radical sign and write the square root of 96 as 4 times the square root of 6. Hope this helps. Good Luck.(11 votes)

- At6:20sal says the sixth root of x^6 is just x. Later (at7:30) he writes the full expression simplified as:

2x * x^(1/3)

My question is: shouldn't it be the absolute value of x instead of x?

2|x| * x^(1/3) instead of 2x * x^(1/3)

What if x is negative? When should I take the absolute value like explained on https://www.khanacademy.org/math/algebra/exponent-equations/simplifying-radical-expressions/v/simplifying-square-roots at3:35?(7 votes)- Yes, I am too. I wonder if x<0, he can still use the exponent formula to do this, because it would let x be just x, not absolute value x. And if we don't know if x<0 or not, we should just use the absolute value. It is quite confusing and I wish he explains it more.(4 votes)

- Hi! I'm in 5 th grade and I still don't get how to do this for most simplifying radicals problems. I have to do this next year in advanced math!!! Help!(2 votes)
- If you have the square root of 52, that's equal to the square root of 4x13. The square root of four is two, but 13 doesn't have a square root that's a whole number. Your answer is 2 (square root of 4) multiplied by the square root of 13.(6 votes)

- I have skill quizes every week in geometry where we are tested on algebra material one problem that always gets me are problems like the following: *∛3x-6=3* how would i solve problems like these or which segment should i look for tutorials on?(3 votes)
- Basically we just need to isolate the missing variable x to the other side of an equation.

Keep in mind though that ∛3x-6=3 and ∛(3x)-6=3 are different from one another

Let me show you how to do it.

If we have ∛3x-6=3 then.......

∛3x-6=3 → This will be our original equation.

∛3x=9 → I added 6 to both sides of the equation

x=9/∛3 → I divided both sides of the equation by ∛3 and there's our answer!

But if what you meant was ∛(3x)-6=3 instead of ∛3x-6=3 then......

∛(3x)-6=3 → This will be our original equation.

∛(3x)=9 → I added 6 to both sides of the equation

3x=9^3 → I raised both sides of the equation by 3

3x=729 → This is what I got

x=243 → I divided both sides of the equation by 3 and there's our answer for this equation.(4 votes)

- at the time2:02in the video, i don't understand what he was doing? Can anyone help me please!(3 votes)
- In this section at2:02, he is finding the 4th root of 16 or ∜16

The 4th root of something is a number multiplied by itself 4 times to equal that something

Well, at that time after factoring 16, he has ∜2∙2∙2∙2 written and he is talking about finding the 4th root of 2 times 2 times 2 times 2

We are looking for a factor that occurs 4 times, and there it is! We can see that 2 is a factor of 16 four different times, so

∜16 = ∜2∙2∙2∙2 = ∜2⁴ = 2

And that is how he got the 2(3 votes)

- In the last example when it was simplified to 2x x^1/3 should the x be |x|?(2 votes)
- Nope.At first it may seem so but no.

We start with (64x^8)^(1/6). Because x is raised to an even power, it will always be non-negative.

We end up with 2x(x)^(1/3). Both x'es are raised to an odd power: therefore if x is negative, so are they. But then we have two negatives, which cancel out: so the second expression is also always positive and fully equal to the first one.(2 votes)

- We know that the sqrt of 9=3 but then is it possible that the sqrt of 9 is -3 since,

-3*-3= +9

pls answer...(2 votes)- Yes, you are correct. Square root of 9 is indeed +3 or -3, which can be written as ±3. In fact any even roots (square root, fourth root, sixth roots, and so on) has two solutions, a positive and a negative. However, when we say "the square root" we often refer to the principal square root, which denotes as √(n). Principal square root refers to the positive solution. We denote it as ±√(n) if we want both positive and negative solutions, or -√(n) for just negative.

Khan has a video about it

https://www.khanacademy.org/math/cc-eighth-grade-math/cc-8th-numbers-operations/cc-8th-roots/v/introduction-to-square-roots(2 votes)

- Why does he keep on using two when finding the prime factorization of it? Do you really have to do it that way?(2 votes)
- The process of prime factorization starts by dividing by the smallest prime number, which is 2. If you can, keep doing it until you can't. At that point look to successively larger primes, 3, 5, 7, 11 . . . .

EG

12 = 2*2*3 since 12/2=6 6/2=3

22 = 2*11 since 22/2=11

21 = 3*7 since 21/3=7

385 = 5*7*11 since 385/5=77 and 77/7=11

As you can see, when you can start on 2, do so, otherwise start with the next smallest prime - this is the convention for prime factorization.

If you choose not to, you will still get the correct answer, but you might also overlook something, so best to follow convention:

24 24/8=6, 6/3=2 so 8*3*2, BUT 8 can still be factored.(2 votes)

## Video transcript

So far, when we were dealing
with radicals we've only been using the square root. We've seen that if I write a
radical sign like this and put a 9 under it, this means the
principal square root of 9, which is positive 3. Or you could view it as the
positive square root of 9. Now, what's implicit when we
write it like this is that I'm taking the square root. So I could have also written
it like this. I could have also written the
radical sign like this and written this index 2 here, which
means the square root, the principal square
root of 9. Find me something that
if I square that something, I get 9. And the radical sign doesn't
just have to apply to a square root. You can change the index here
and then take an arbitrary root of a number. So for example, if I were to
ask you, what-- You could imagine this is called the cube
root, or you could call it the third root of 27. What is this? Well, this is some number that
if I take it to the third power, I'd get 27. Well, the only number that if
you take it to the third power, you get 27
is equal to 3. 3 times 3 times 3
is equal to 27. 9 times 3, 27. So likewise, let me
just do one more. So if I have 16-- I'll do
it in a different color. If I have 16 and I want to take
the fourth root of 16, what number times itself
4 times is equal to 16? And if it doesn't pop out at
you immediately, you can actually just do a prime
factorization of 16 to figure it out. Let's see. 16 is 2 times 8. 8 is 2 times 4. 4 is 2 times 2. So this is equal to the fourth
root of 2 times 2 times 2 times 2. You have these four 2's here. Well, I have four 2's being
multiplied, so the fourth root of this must be equal to 2. And you could also view this
as kind of the fourth principal root because if these
were all negative 2's, it would also work. Just like you have multiple
square roots, you have multiple fourth roots. But the radical sign implies
the principal root. Now, with that said, we've
simplified traditional square roots before. Now we should hopefully be able
to simplify radicals with higher power roots. So let's try a couple. Let's say I want to simplify
this expression. The fifth root of 96. So like I said before, let's
just factor this right here. So 96 is 2 times 48. Which is 2 times 24. Which is 2 times 12. Which is 2 times 6. Which is 2 times 3. So this is equal to the fifth
root of 2 times 2 times 2 times 2 times 2. Times 3. Or another way you could view
it, is you could view it to a fractional power. You could view it to
a fractional power. We've talked about
that already. This is the same thing as 2
times 2 times 2 times 2 times 2 times 3 to the 1/5 power. Let me make this clear. Having an nth root of some
number is equivalent to taking that number to the 1/n power. These are equivalent statements
right here. So if you're taking this to
the 1/5 power, this is the same thing as taking 2 times
2 times 2 times 2 times 2 to the 1/5. Times 3 to the 1/5. Now I have something
that's multiplied. I have 2 multiplied
by itself 5 times. And I'm taking that
to the 1/5. Well, the 1/5 power of this
is going to be 2. Or the fifth root of this
is just going to be 2. So this is going to
be a 2 right here. And this is going to be
3 to the 1/5 power. 2 times 3 to the 1/5, which is
this simplified about as much as you can simplify it. But if we want to keep in
radical form, we could write it as 2 times the fifth
root 3 just like that. Let's try another one. Let me put some variables
in there. Let's say we wanted to simplify
the sixth root of 64 times x to the eighth. So let's do 64 first. 64 is equal to 2 times 32,
which is 2 times 16. Which is 2 times 8. Which is 2 times 4. Which is 2 times 2. So we have 1, 2, 3, 4, 5, 6. So it's essentially 2
to the sixth power. So this is equivalent to the
sixth root of 2 to the sixth-- that's what 64 is --times
x to the eighth power. Now, the sixth root of 2 to
the sixth, that's pretty straightforward. So this part right here is just
going to be equal to 2. That's going to be 2 times
the sixth root of x to the eighth power. And how can we simplify this? Well, x to the eighth power,
that's the same thing as x to the sixth power times
x squared. You have the same base, you
would add the exponents. This is the same thing
as x to the eighth. So this is going to be equal to
2 times the sixth root of x to the sixth times x squared. And the sixth root, this part
right here, the sixth root of x to the sixth, that's just x. So this is going to be equal to
2 times x times the sixth root of x squared. Now, we can simplify this
even more if you really think about. Remember, this expression right
here, this is the exact same thing as x squared
to the 1/6 power. And if you remember your
exponent properties, when you raise something to an exponent,
and then raise that to an exponent, that's
equivalent to x to the 2 times 1/6 power. Or-- let me write this --2 times
1/6 power, which is the same thing-- Let me not forget
to write my 2x there. So I have a 2x there
and a 2x there. And this is the same thing as
2x-- it's the same 2x there --times x to the 2/6. Or, if we want to write that in
most simple form or lowest common form, you get 2x times
x to the-- What do you have here? x to the 1/3. So if you want to write it in
radical form, you could write this is equal to 2 times 2x
times the third root of x. Or, the other way to think about
it, you could just say-- So we could just go from
this point right here. We could write this. We could ignore this,
what we did before. And we could say, this is the
same thing as 2 times x to the eighth to the 1/6 power. x to the eighth to
the 1/6 power. So this is equal to 2
times x to the-- 8 times 1/6 --8/6 power. Now we can reduce
that fraction. That's going to be 2 times
x to the 4/3 power. And this and this are completely
equivalent. Why is that? Because we have 2 times x or 2
times x to the first power times x to the 1/3 power. You add 1 to 1/3, you get 4/3. So hopefully you found this
little tutorial on higher power radicals interesting. And I think it is useful to kind
of see it in prime factor form and realize, oh, if I'm
taking the sixth root, I have to find a prime factor that
shows up at least six times. And then I could figure out
that's 2 to the sixth. Anyway, hopefully you found
this mildly useful.