Question

The function $\frac{2(\sqrt{2}+\sqrt{6})}{3\left(\sqrt{2+\sqrt{3}}\right)}$ is equal to

• A. $\frac{2\sqrt{2}}{3}$
• B. $1$
• C. $\frac{2\sqrt{3}}{3}$
• D. $\frac{4}{3}$

Answer 1

The correct option is A $\frac{2\sqrt{2}}{3}$

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

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

squaring both sides,

$9x^2(2+\sqrt{3})=4(\sqrt{2}+\sqrt{6}{)}^2=4(8+4\sqrt{3})$

$9x^2(2+\sqrt{3})=4.4(2+\sqrt{3})$

$x^2=\frac{16}{9}\Rightarrow x=\pm \frac{4}{3}$

$\because x\ne \frac{-4}{3}$, so $x=4/3$