How do you simplify $\sqrt{49 m^{2} n^{8}}$ $sqrt (49 m^2n^ 8)$ ?

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Answer 1 · 2018-03-27T12:18:36

$7 m n^{4}$ $7mn^4$

First, we know that $a \sqrt{b} c$ $asqrtbc$ is the same as $a \sqrt{b} \cdot a \sqrt{c}$ $asqrtb*asqrtc$

Therefore, $\sqrt{49} m^{2} n^{8}$ $sqrt49m^2n^8$ is the same as $\sqrt{49} \cdot {\sqrt{m}}^{2} \cdot {\sqrt{n}}^{8}$ $sqrt49*sqrtm^2*sqrtn^8$

And we know that $\sqrt{49}$ $sqrt49$ is $7$ $7$ so we would have
$7 {\sqrt{m}}^{2} \cdot {\sqrt{n}}^{8}$ $7sqrtm^2*sqrtn^8$

We also know that $a {\sqrt{b}}^{a}$ $asqrtb^a$ is the same as $b$ $b$ , so ${\sqrt{m}}^{2}$ $sqrtm^2$ is $m$ $m$ .

We would now have $7 m {\sqrt{n}}^{8}$ $7msqrtn^8$ .

We will apply the rule that $a {\sqrt{b}}^{c} = b^{\frac{c}{a}}$ $asqrtb^c=b^(c/a)$

So, $7 m {\sqrt{n}}^{8} = 7 m n^{\frac{8}{2}}$ $7msqrtn^8=7mn^(8/2)$ and $\frac{8}{2}$ $8/2$ is $4$ $4$

Final answer: $7 m n^{4}$ $7mn^4$ .

How do you simplify sqrt (49 m^2n^ 8)√49m2n8sqrt (49 m^2n^ 8)?