How do you add $\sqrt{24} + \sqrt{54}$ $sqrt24 + sqrt54$ ?

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Answer 1 · 2015-04-04T17:50:30

The answer is $5 \sqrt{6}$ $5sqrt(6)$

In this type of problems, there is a rule zero: Prime Factorization.

$24 = 2^{3} \cdot 3$ $24=2^3 *3$ and $54 = 3^{3} \cdot 2$ $54 = 3^3 *2$

So

$\sqrt{24} + \sqrt{54} = \sqrt{2^{3} \cdot 3} + \sqrt{3^{3} \cdot 2}$ $sqrt(24) + sqrt(54) = sqrt(2^3*3) + sqrt(3^3*2)$

$= \sqrt{2^{2} \cdot 2 \cdot 3} + \sqrt{3^{2} \cdot 3 \cdot 2}$ $= sqrt(2^2*2*3) + sqrt(3^2*3*2)$

$= 2 \sqrt{2 \cdot 3} + 3 \sqrt{3 \cdot 2}$ $=2sqrt(2*3)+3sqrt(3*2)$

$2 \sqrt{6} + 3 \sqrt{6} = \sqrt{6} (2 + 3) = 5 \sqrt{6}$ $2sqrt(6)+3sqrt(6) = sqrt(6)(2+3) = 5sqrt(6)$

How do you add √24+√54sqrt24 + sqrt54?