Question

Simplify the following:
$\frac{\sqrt{21}}{\sqrt{3}+\sqrt{7}}+\frac{2\sqrt{5}}{\sqrt{21}+\sqrt{5}}$.

Answer 1

We rationalise the denominators of $\left(\frac{\sqrt{21}}{\sqrt{3}+\sqrt{7}}\right)+\left(\frac{2\sqrt{5}}{\sqrt{21}+\sqrt{5}}\right)$ and then simplify as follows:

$\left(\frac{\sqrt{21}}{\sqrt{3}+\sqrt{7}}\right)+\left(\frac{2\sqrt{5}}{\sqrt{21}+\sqrt{5}}\right)=(\frac{\sqrt{21}}{\sqrt{3}+\sqrt{7}}\times \frac{\sqrt{3}-\sqrt{7}}{\sqrt{3}-\sqrt{7}})-(\frac{2\sqrt{5}}{\sqrt{21}+\sqrt{5}}\times \frac{\sqrt{21}-\sqrt{5}}{\sqrt{21}-\sqrt{5}})$

$=\left(\frac{\sqrt{21}(\sqrt{3}-\sqrt{7})}{(\sqrt{3}+\sqrt{7})(\sqrt{3}-\sqrt{7})}\right)-\left(\frac{2\sqrt{5}(\sqrt{21}-\sqrt{5})}{(\sqrt{21}+\sqrt{5})(\sqrt{21}-\sqrt{5})}\right)$

$=\left(\frac{\sqrt{63}-\sqrt{147}}{(\sqrt{3}{)}^2-(\sqrt{7}{)}^2}\right)-\left(\frac{2\sqrt{105}-2\sqrt{25}}{(\sqrt{21}{)}^2-(\sqrt{5}{)}^2}\right)$

$=\frac{-(\sqrt{147}-\sqrt{63})}{-(7-3)}-\frac{2(\sqrt{105}-5)}{21-5}=\frac{\sqrt{49\times 3}-\sqrt{9\times 7}}{4}+\frac{\sqrt{105}-5}{8}=\frac{7\sqrt{3}-3\sqrt{7}}{4}+\frac{\sqrt{105}-5}{8}$

$=\frac{2(7\sqrt{3}-3\sqrt{7})+\sqrt{105}-5}{8}=\frac{14\sqrt{3}-6\sqrt{7}+\sqrt{105}-5}{8}$ .

Hence, $\left(\frac{\sqrt{21}}{\sqrt{3}+\sqrt{7}}\right)+\left(\frac{2\sqrt{5}}{\sqrt{21}+\sqrt{5}}\right)=\frac{14\sqrt{3}-6\sqrt{7}+\sqrt{105}-5}{8}$ .