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

Jul 17, 2016

$5 y , - 5 y$

#### Explanation:

Anything to the $\frac{1}{2}$ power is the same as the square root of that value.

${\left(25 {y}^{2}\right)}^{\frac{1}{2}} = \sqrt{25 {y}^{2}}$

Find the square root. Note that $\sqrt{a b} = \sqrt{a} \times \sqrt{b}$

$\sqrt{25} \times \sqrt{{y}^{2}}$

$5 \times 5$ is $25$, to the square root of $25$ is $5$. Using the rule from the very top, $\sqrt{{y}^{2}} = {\left({y}^{2}\right)}^{\frac{1}{2}} = {y}^{2 \cdot \frac{1}{2}} = {y}^{1} = y$.

$5 \times y$

$5 y$

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: $\sqrt{16} = 4 , - 4$ since $4 \cdot 4 = 16$ and $- 4 \cdot - 4 = 16$.

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

The other possible solution: $- 5 y$