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## Algebra (all content)

### 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! •   its plus and minus 10

+10 and -10
they both give you 100
• 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? • Is there a faster way to do prime factorization • 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.
• At sal says the sixth root of x^6 is just x. Later (at ) 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 at ? • 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! • 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? • 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.
• at the time in the video, i don't understand what he was doing? Can anyone help me please! • In this section at , 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
• In the last example when it was simplified to 2x x^1/3 should the x be |x|? • 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.
• We know that the sqrt of 9=3 but then is it possible that the sqrt of 9 is -3 since,
-3*-3= +9 • Why does he keep on using two when finding the prime factorization of it? Do you really have to do it that way? • 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.