How do you simplify $\sqrt{32} \cdot \sqrt{144}$ $sqrt32*sqrt144$ ?

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How do you simplify $\sqrt{32} \cdot \sqrt{144}$ $sqrt32*sqrt144$ ?

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Answer 1 · 2016-07-01T11:00:17

$48 \sqrt{2} .$ $48sqrt2.$

Explanation:

We use prime factorisation of $32 = 2^{5}$ $32=2^5$ and $144 = 2^{4} \cdot 3^{2} .$ $144=2^4*3^2.$

Hence the expression, $= \sqrt{32} \cdot \sqrt{144},$ $=sqrt32*sqrt144,$
$= {(2^{5})}^{\frac{1}{2}} \cdot {(2^{4} \cdot 3^{2})}^{\frac{1}{2}}$ $=(2^5)^(1/2)*(2^4*3^2)^(1/2)$ = ${(2^{4} \cdot 2)}^{\frac{1}{2}} \cdot {(2^{4} \cdot 3^{2})}^{\frac{1}{2}}$ $(2^4*2)^(1/2)*(2^4*3^2)^(1/2)$
$= {{(2^{4})}^{\frac{1}{2}} 2^{\frac{1}{2}}} {{(2^{4})}^{\frac{1}{2}} \cdot {(3^{2})}^{\frac{1}{2}}}$ $={(2^4)^(1/2)2^(1/2)}{(2^4)^(1/2)*(3^2)^(1/2)}$ = $2^{4 \cdot \frac{1}{2}} \cdot 2^{\frac{1}{2}} \cdot 2^{4 \cdot \frac{1}{2}} \cdot 3^{2 \cdot \frac{1}{2}} = 2^{2} \cdot 2^{\frac{1}{2}} \cdot 2^{2} \cdot 3^{1} = 4 \cdot \sqrt{2} \cdot 4 \cdot 3 = 48 \sqrt{2} .$ $2^(4*1/2)*2^(1/2)*2^(4*1/2)*3^(2*1/2)=2^2*2^(1/2)*2^2*3^1=4*sqrt2*4*3=48sqrt2.$

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How do you simplify $\sqrt{32} \cdot \sqrt{144}$ $sqrt32*sqrt144$ ?

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How do you simplify $\sqrt{32} \cdot \sqrt{144}$ $sqrt32*sqrt144$ ?