How do you solve ${(25 y^{2})}^{\frac{1}{2}}$ $(25y^2)^(1/2)$ ?

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How do you solve ${(25 y^{2})}^{\frac{1}{2}}$ $(25y^2)^(1/2)$ ?

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Answer 1 · 2016-07-17T15:51:35

$5 y, - 5 y$ $5y, -5y$

Explanation:

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

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

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

$\sqrt{25} \times \sqrt{y^{2}}$ $sqrt25 times sqrt(y^2)$

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

$5 \times y$ $5 times y$

$5 y$ $5y$

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: $\sqrt{16} = 4, - 4$ $sqrt16 = 4, -4$ since $4 \cdot 4 = 16$ $4*4 = 16$ and $- 4 \cdot - 4 = 16$ $-4*-4 = 16$ .

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

The other possible solution: $- 5 y$ $-5y$

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