Question

How do you solve `(25y^2)^(1/2)`?

Answer 1

`5y, -5y`

Explanation:

Anything to the `1/2` power is the same as the square root of that value.

`(25y^2)^(1/2) = sqrt(25y^2)`

Find the square root. Note that `sqrt(ab) = sqrtaxxsqrtb`

`sqrt25 times sqrt(y^2)`

`5 times 5` is `25`, to the square root of `25` is `5`. Using the rule from the very top, `sqrt(y^2) = (y^2)^(1/2) = y^(2*1/2) = y^1 = y`.

`5 times y`

`5y`

There can also be a negative answer. Every value has a negative root, because when you multiply two negative numbers you end up with the same positive number.

Example: `sqrt16 = 4, -4` since `4*4 = 16` and `-4*-4 = 16`.

So, the square root of `25` would actually be `5` and `-5`.

The other possible solution: `-5y`