Question

If a and b are rational numbers and $\frac{\sqrt{11}-\sqrt{7}}{\sqrt{11}+\sqrt{7}}=a-b\sqrt{77},$ find $a-b$.

• A. 4

Answer 1

The correct option is A 4
We have
$\frac{\sqrt{11}-\sqrt{7}}{\sqrt{11}+\sqrt{7}}=\frac{(\sqrt{11}-\sqrt{7})}{(\sqrt{11)}+\sqrt{7}}\times \frac{(\sqrt{11}-\sqrt{7})}{(\sqrt{11}-\sqrt{7})}=\frac{(\sqrt{11}-\sqrt{7}{)}^2}{\left\{\right(\sqrt{11}{)}^2-\left(\sqrt{7}{)}^2\right\}}=\frac{(11+7-2\times \sqrt{1}1\times \sqrt{7})}{(11-7)}=\frac{18-2\sqrt{77}}{4}=(\frac{18}{4}-\frac{2}{4}\sqrt{77})=(\frac{9}{2}-\frac{1}{2}.\sqrt{77})\therefore \frac{\sqrt{11}-\sqrt{7}}{\sqrt{11}+\sqrt{7}}=a-b\sqrt{77}\Rightarrow \frac{9}{2}-\frac{1}{2}.\sqrt{77}=a-b\sqrt{77}\Rightarrow a=\frac{9}{2}andb=\frac{1}{2}$
Hence~ $a=\frac{9}{2}andb=\frac{1}{2}$
$a-b=\frac{9}{2}-\frac{1}{2}=\frac{8}{2}=4$