Question

Rationalize the denominator of the expression $\frac{4}{\sqrt{8}-\sqrt{2}}.$

• A. $\frac{2(\sqrt{8}+\sqrt{2})}{3}$
• B. $\frac{3(\sqrt{8}+\sqrt{2})}{2}$
• C. $\frac{2}{3(\sqrt{8}+\sqrt{2})}$

Answer 1

The correct option is A $\frac{2(\sqrt{8}+\sqrt{2})}{3}$

To rationalize subtaction of irrational numbers, apply $a^2-b^2=(a+b)(a-b).$
Here, $a=\sqrt{8}$ and $b=\sqrt{2}$

Multiply the expression by $\sqrt{8}+\sqrt{2}$ in numerator and denominator.

$\frac{4}{\sqrt{8}-\sqrt{2}}$
$=\frac{4}{\sqrt{8}-\sqrt{2}}\times \frac{\sqrt{8}+\sqrt{2}}{\sqrt{8}+\sqrt{2}}$
$=\frac{4\times (\sqrt{8}+\sqrt{2})}{(\sqrt{8}-\sqrt{2})(\sqrt{8}+\sqrt{2})}$
$=\frac{4\times (\sqrt{8}+\sqrt{2})}{(\sqrt{8}{)}^2-(\sqrt{2}{)}^2}$
$=\frac{4\times (\sqrt{8}+\sqrt{2})}{8-2}$
$=\frac{4\times (\sqrt{8}+\sqrt{2})}{6}$
$=\frac{2(\sqrt{8}+\sqrt{2})}{3}$
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