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## Comment on

Other Roots## Do we need to practice nth

Does GRE expects us to know higher square root of numbers?

## You certainly won't have to

You certainly won't have to EVALUATE/CALCULATE any nth roots where n > 10.

However, it's conceivable that you might have to SIMPLIFY an expression in which there's an nth root where n > 10

For example, we can simplify the 20th root of x^100 as follows:

20th root of x^100 = (x^100)^(1/20)

= x^(100/20)

= x^5

## Could you go through the

I keep getting x^3 when the answer is d) x^5.

## GIVEN: x(x²)³/x²

GIVEN: x(x²)³/x²

First we deal with (x²)³ [BEDMAS/PEMDAS says to deal with the exponent first]

Apply Power of Power Law to get: x(x⁶)/x²

Rewrite numerator as: x¹(x⁶)/x²

Apply Product Law to numerator to get : x⁷/x²

Apply Quotient Law to get: x⁵

Cheers,

Brent

## any number with a negative

## I'm not certain what you are

I'm not certain what you are asking.

On the GRE, we say that √(-1) is not a real number.

In fact, the SQUARE ROOT of any negative number is not a real number.

HOWEVER, there are other roots in which the root is a real number.

For example, ∛(-8) = -2, since (-2)³ = -8

Does that help?

Cheers,

Brent

## Hi Brent,

If n<0, square root(n) is not defined. But, the cuberoot(n), where n is -125 is having a value of 5?

I am not able to comprehend.

Thanks!

Ketan

## Here's a little background

Here's a little background about square roots:

√9 = 3, since 3² = 9

√25 = 5, since 5² = 25

To evaluate √(-16), we'd have to find a value so that, when SQUARED, we get -16

In other words, we want: (something)² = -16

Since no real number can satisfy the above requirement, we say that √(-16) is not defined.

------------------

A little background about square roots:

∛8 = 2, since 2³ = 8

∛64 = 4, since 4³ = 64

To evaluate ∛(-125), we need a value so that, when we CUBED, we get -125

In other words, we want: (something)³ = -125

In this case, there is a value that satisfies the above requirement.

We know that (-5)(-5)(-5) = -125

In other words, (-5)³ = -125

As such, we can say that ∛(-125) = -5

Likewise, ∛(-8) = -2, and ∛(-27) = -3

Does that help?

## Hi Brent,

https://greprepclub.com/forum/what-is-the-value-of-approximated-to-the-nearest-integer-9069.html

In this question, shouldn't the answer be 'C'. Since the value is still greater than 4?

Thanks!

Ketan

## Question link: https:/

Question link: https://greprepclub.com/forum/what-is-the-value-of-approximated-to-the-n...

Our goal is to approximate the expression to the nearest INTEGER.

The when we try to evaluate the expression, we see that it's between 4 and 5

So, the correct answer must be C or D.

In my solution, I show that the 5th root of 34 is very very close to 2.

So, when we double that amount, the result will be closer to 4 than it is to 5.

Since it's closer to 4 than it is to 5, the best answer is D.

## Hi Brent,

In one of the question I saw squareroot(12.5)+ squareroot(12.5) so why shuld we add them because they are the same digits right.

Thanks

## Do you have a link to the

Do you have a link to the question?

In general, we know that k + k = 2k

And x³ + x³ = 2x³

And 5w + 5w = 2(5w) = 10w

etc.

Likewise, √12.5 + √12.5 = 2√12.5

And √3 + √3 = 2√3

Does that help?

Cheers,

Brent