What is (square root of [6] + 2 square root of [2])(4square root of [6] - 3 square root of 2)?

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What is (square root of [6] + 2 square root of [2])(4square root of [6] - 3 square root of 2)?

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Answer 1 · 2015-09-17T03:35:46

$12 + 5 \sqrt{12}$ $12 + 5sqrt12$

Explanation:

We multiply cross-multiply, that is,
$(\sqrt{6} + 2 \sqrt{2}) (4 \sqrt{6} - 3 \sqrt{2})$ $(sqrt6 + 2sqrt2)(4sqrt6 - 3sqrt2)$
equals
$\sqrt{6} \cdot 4 \sqrt{6} + 2 \sqrt{2} \cdot 4 \sqrt{6} - \sqrt{6} \cdot 3 \sqrt{2} - 2 \sqrt{2} \cdot 3 \sqrt{2}$ $sqrt6*4sqrt6 + 2sqrt2*4sqrt6 -sqrt6*3sqrt2 - 2sqrt2*3sqrt2$

The square roots time themselves equal the number under the root, so
$4 \cdot 6 + 8 \sqrt{2} \sqrt{6} - 3 \sqrt{6} \sqrt{2} - 6 \cdot 2$ $4*6 +8sqrt2sqrt6 - 3sqrt6sqrt2 - 6*2$

We put $\sqrt{2} \sqrt{6}$ $sqrt2sqrt6$ in evidence:
$24 + (8 - 3) \sqrt{6} \sqrt{2} - 12$ $24 + (8-3)sqrt6sqrt2 - 12$

We can join these two roots in one, after all $\sqrt{x} \sqrt{y} = \sqrt{x y}$ $sqrtxsqrty = sqrt(xy)$ as long as they're not both negative. So, we get
$24 + 5 \sqrt{12} - 12$ $24 + 5sqrt12 - 12$

Finally, we just take the difference of the two constants and call it a day
$12 + 5 \sqrt{12}$ $12 + 5sqrt12$

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What is (square root of [6] + 2 square root of [2])(4square root of [6] - 3 square root of 2)?

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