Question

How do you simpilify `sqrt(125x^2)`?

Answer 1

`sqrt(125x^2)=5sqrt(5)x`

Explanation:

A square root can also be seen as a power of `1//2`, so we can rewrite our expression as

`(125x^2)^(1/2)`

In this case, the power is distributed across the factors, which is the same as saying that the square-root can be separated across the factors as follows

`125^(1/2) * (x^2)^(1/2) implies sqrt(125)*sqrt(x^2)`

In the second term, the powers simply multiply

`125^(1/2) * x^(2*1/2) = 125^(1/2) * x implies sqrt(125) * x`

Which gives us the rather un-surprising result that the square-root of `x`-squared is simply `x`. We can look at this as a process of looking for paired factors under the square root and pulling one of these to the outside, i.e. `sqrt(x^2)=sqrt(ulx*ulx)=x`. Let's use this method on the last factor

`sqrt(125)*x = sqrt(5*ul5*ul5)*x=5sqrt(5)x`

Therefore

`sqrt(125x^2)=5sqrt(5)x`

Answer 2

`" "5xsqrt(5)`

Explanation:

Let us determine if there are any squared values in the prime factors of 125

We have one value that we can take the square root of`->5^2`
as `5xx5xx5-> 5xx5^2`

Write as:`" "sqrt(5xx5^2xx x^2)`

Taking the `5^2" and "x^2` outside of the root

`" "5xsqrt(5)`