Question

The expression $\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}$ when simplified reduces to $a-b\sqrt{15}$.

Then find the value of a + b.

Answer 1

In $\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}$, the denominator has to be rationalised for simplification. So,

$\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}=\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}$$\times$$\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}-\sqrt{3}}$$\left[1mark\right]$

$=\frac{(\sqrt{5}-\sqrt{3}{)}^2}{{\sqrt{5}}^2-{\sqrt{3}}^2}$

$=\frac{1}{2}[{\left(\sqrt{5}\right)}^2+{\left(\sqrt{3}\right)}^2-2\sqrt{15}]$

$=\frac{1}{2}[5+3-2\sqrt{15}]$

$=4-\sqrt{15}$
$\text{So, a = 4 and b = 1}$
$\therefore$ a + b = 4 + 1 = 5$\left[1mark\right]$