Question

Evaluate $\dfrac{2\sqrt{72}}{\sqrt{8}+\sqrt{2}}$.

Answer 1

4

Explanation:

We first simplify $\sqrt{72}$ and $\sqrt{8}$:\begin{align*}

\sqrt{72} &=\sqrt{36\cdot 2} = \sqrt{36}\cdot \sqrt{2} = 6\sqrt{2},\\

\sqrt{8} &= \sqrt{4\cdot 2} = \sqrt{4}\cdot \sqrt{2} = 2\sqrt{2}.

\end{align*}Then, we have

\[\dfrac{2\sqrt{72}}{\sqrt{8}+\sqrt{2}} = \dfrac{2\left(6\sqrt{2}\right)}{2\sqrt{2} + \sqrt{2}}

= \dfrac{12\sqrt{2}}{3\sqrt{2}} = \dfrac{12}{3}\cdot \dfrac{\sqrt{2}}{\sqrt{2}} = \boxed{4}.\]

Answer 2

Answer: 4

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Work Shown:

`(2√(72))/(√(8)+√(2))\\\\(2√(36*2))/(√(4*2)+√(2))\\\\(2√(36)*√(2))/(√(4)*√(2)+√(2))\\\\(2*6*√(2))/(2*√(2)+√(2))\\\\(12√(2))/(2√(2)+√(2))\\\\(12√(2))/(3√(2))\\\\(12)/(3)\\\\4`

Note in step 2, I factored each number in the square root to pull out the largest perfect square factor. From there, I used the rule that
`√(A*B) = √(A)*√(B)` to break up the roots.