Question #dcd3f

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Question #dcd3f

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Answer 1 · 2017-05-08T19:38:34

$13 \sqrt{5}$ $13sqrt5$

Explanation:

First, simplify the radicals:
$4 \sqrt{4 \times 5}$ $4sqrt(4 xx 5)$ $+$ $+$ $3 \sqrt{9 \times 5}$ $3sqrt(9 xx 5)$ $-$ $-$ $2 \sqrt{4 \times 5}$ $2sqrt(4 xx 5)$

Next, take out the perfect squares (which are $4$ $4$ and $9$ $9$ ):
$4 (2) \sqrt{5}$ $4(2)sqrt5$ $+$ $+$ $3 (3) \sqrt{5}$ $3(3)sqrt5$ $-$ $-$ $2 (2) \sqrt{5}$ $2(2)sqrt5$

Multiply:
$8 \sqrt{5}$ $8sqrt5$ $+$ $+$ $9 \sqrt{5}$ $9sqrt5$ $-$ $-$ $4 \sqrt{5}$ $4sqrt5$

Add and subtract:
$13 \sqrt{5}$ $13sqrt5$

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