How do you divide $5 \sqrt{16 y^{4}} + 7 \sqrt{2 y}$ $5sqrt(16y^4) + 7sqrt(2y)$ ?

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How do you divide $5 \sqrt{16 y^{4}} + 7 \sqrt{2 y}$ $5sqrt(16y^4) + 7sqrt(2y)$ ?

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Answer 1 · 2018-08-06T03:59:13

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$5 \sqrt{16 y^{4}} + 7 \sqrt{2 y} = 20 y^{2} + 7 \sqrt{2} \sqrt{y}$ $color(red)(5sqrt(16y^4) + 7sqrt(2y)=20y^2+7sqrt(2)sqrt(y)$

Explanation:

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Given: $5 \sqrt{16 y^{4}} + 7 \sqrt{2 y}$ $color(red)(5sqrt(16y^4) + 7sqrt(2y)$

My understanding of the question: Simplify the radical expression.

$\Rightarrow (5) (\sqrt{16}) \sqrt{y^{4}} + 7 \sqrt{2} \sqrt{y}$ $rArr (5)(sqrt(16))sqrt(y^4) + 7sqrt(2)sqrt(y)$

$\Rightarrow 5 \cdot 4 \cdot y^{2} + 7 \sqrt{2} \sqrt{y}$ $rArr 5*4*y^2+7sqrt(2)sqrt(y)$

$\Rightarrow 20 y^{2} + 7 \sqrt{2} \sqrt{y}$ $rArr 20y^2+7sqrt(2)sqrt(y)$

You can stop here.

If you so wish, you can simplify in a different way.

Consider the step:

$\Rightarrow (5) (\sqrt{16}) \sqrt{y^{4}} + 7 \sqrt{2} \sqrt{y}$ $rArr (5)(sqrt(16))sqrt(y^4) + 7sqrt(2)sqrt(y)$

$\Rightarrow (5) (\sqrt{16}) \sqrt{y^{3}} \sqrt{y} + 7 \sqrt{2} \sqrt{y}$ $rArr (5)(sqrt(16))sqrt(y^3)sqrt(y) + 7sqrt(2)sqrt(y)$

$\Rightarrow 5 \cdot 4 \cdot \sqrt{y^{3}} \sqrt{y} + 7 \sqrt{2} \sqrt{y}$ $rArr 5*4*sqrt(y^3)sqrt(y)+7sqrt(2) sqrt(y)$

$\Rightarrow \sqrt{y} [20 \sqrt{y^{3}} + 7 \sqrt{2}]$ $rArr sqrt(y)[20sqrt(y^3)+7sqrt(2)]$

Hope this helps.

Answer 2 · 2018-08-13T12:27:57

$\Rightarrow (\frac{40}{7}) {(y)}^{\frac{3}{2}}$ $color(magenta)(=> (40/7) (y)^(3/2)$

Explanation:

I am taking the sum as

$\frac{5 \cdot \sqrt{16 y^{4}}}{7 \sqrt{2 y}}$ $(5 * sqrt (16y^4)) / (7 sqrt (2y))$

as it has been asked to divide.

$\Rightarrow \frac{5 \cdot \sqrt{2^{4} y^{4}}}{7 \sqrt{2 y}}$ $=> (5 * sqrt (2^4y^4)) / ( 7 sqrt (2y)$

$\Rightarrow (\frac{5}{7}) \cdot \frac{\sqrt{2^{4} y^{4}}}{\sqrt{2} y}$ $=> (5/7) * sqrt (2^4y^4)/(sqrt 2y)$

$\Rightarrow (\frac{5}{7}) \cdot \sqrt{2^{3} y^{3}}$ $=> (5/7) * sqrt (2^3y^3)$

$\Rightarrow (\frac{40}{7}) {(y)}^{\frac{3}{2}}$ $=> (40/7) (y)^(3/2)$

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