# Lesson: Other Roots

## Comment on Other Roots

### Do we need to practice nth

Do we need to practice nth root of a number where n>10.
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

Could you go through the steps to solve 151(6) from the Official guide 2nd edition?

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

any number with a negative sign under square root isnt a complex number and isn't it iyota (i=(-1)^1/n? ### 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,

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,

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 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,

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