Question

$\frac{\left(\sqrt{11}-\sqrt{7}\right)}{\left(\sqrt{11}+\sqrt{7}\right)}=a-\sqrt{77}b$. Find $a$ and $b.$

Answer 1

Step 1: Rationalizing the denominator

$$\begin{gathered} \frac{\left(\sqrt{11}-\sqrt{7}\right)}{\left(\sqrt{11}+\sqrt{7}\right)}=\frac{\left(\sqrt{11}-\sqrt{7}\right)}{\left(\sqrt{11}+\sqrt{7}\right)}\times \frac{\left(\sqrt{11}-\sqrt{7}\right)}{\left(\sqrt{11}-\sqrt{7}\right)} \\ =\frac{{(\sqrt{11}-\sqrt{7})}^2}{{\left(\sqrt{11}\right)}^2-{\left(\sqrt{7}\right)}^2}\left[\because \left(a-b\right)\left(a+b\right)=a^2-b^2\right] \\ =\frac{{\left(\sqrt{11}\right)}^2+{\left(\sqrt{7}\right)}^2-2\sqrt{11}\sqrt{7}}{11-7}\left[\because {\left(a-b\right)}^2=a^2-2ab+b^2\right] \\ =\frac{11+7-2\sqrt{77}}{4} \\ =\frac{18-2\sqrt{77}}{4} \\ =\frac{18}{4}-\frac{2\sqrt{77}}{4} \\ \mathbf{=}\frac{\mathbf{9}}{\mathbf{2}}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}\sqrt{\mathbf{77}} \end{gathered}$$

Step 2: Compare the rational and irrational part

Now,

$a-\sqrt{77}b=\frac{9}{2}-\frac{1}{2}\sqrt{77}$

Thus,

$\mathbf{\therefore }\mathbf{}\mathit{a}\mathbf{=}\frac{\mathbf{9}}{\mathbf{2}}$ and $\mathit{b}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}$