How do you simplify $\sqrt{32} + \sqrt{8}$ $sqrt32 + sqrt8$ ?

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How do you simplify $\sqrt{32} + \sqrt{8}$ $sqrt32 + sqrt8$ ?

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Answer 1 · 2015-05-18T13:36:40

We can decouple the numbers inside the square roots:

$\sqrt{2 \cdot 2 \cdot 2 \cdot 2 \cdot 2} + \sqrt{2 \cdot 2 \cdot}$ $sqrt(2*2*2*2*2)+sqrt(2*2*)$

As we are dealing with square roots, we can factor out all the squared numbers from inside the roots:

$\sqrt{2^{2} \cdot 2^{2} \cdot 2} + \sqrt{2^{2} \cdot 2}$ $sqrt(2^2*2^2*2)+sqrt(2^2*2)$
$\sqrt{4 \cdot 4 \cdot 2} + \sqrt{4 \cdot 2}$ $sqrt(4*4*2)+sqrt(4*2)$
$\sqrt{16 \cdot 2} + \sqrt{4 \cdot 2}$ $sqrt(16*2)+sqrt(4*2)$
$4 \sqrt{2} + 2 \sqrt{2}$ $4sqrt(2)+2sqrt(2)$
$6 \sqrt{2}$ $6sqrt(2)$

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How do you simplify $\sqrt{32} + \sqrt{8}$ $sqrt32 + sqrt8$ ?

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How do you simplify $\sqrt{32} + \sqrt{8}$ $sqrt32 + sqrt8$ ?