What is $\sqrt{8} + \sqrt{18} - \sqrt{32}$ $sqrt(8)+sqrt(18)-sqrt(32)$ ?

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Answer 1 · 2018-03-11T13:02:30

$9 \sqrt{2}$ $9sqrt2$

Before we start, do you notice something? The numbers $8$ $8$ , $18$ $18$ , and $32$ $32$ are perfect squares multiplied by $2$ $2$ .

$\sqrt{8} + \sqrt{18} - \sqrt{32}$ $sqrt8+sqrt18-sqrt32$

Split up the invidual square roots,

$(\sqrt{4} \times \sqrt{2}) + (\sqrt{9} \times \sqrt{2}) + (\sqrt{16} \times \sqrt{2})$ $(sqrt4xxsqrt2)+(sqrt9xxsqrt2)+(sqrt16xxsqrt2)$

Square root the perfect squares,

$2 \sqrt{2} + 3 \sqrt{2} + 4 \sqrt{2}$ $2sqrt2+3sqrt2+4sqrt2$

Add them all up,

$9 \sqrt{2}$ $9sqrt2$

Done :D

What is √8+√18−√32sqrt(8)+sqrt(18)-sqrt(32)?