Question

If $p$ =$\frac{7-\sqrt{11}}{7+\sqrt{11}}$ and $q$ =$\frac{7+\sqrt{11}}{7-\sqrt{11}}$,then find the value of $(p+q{)}^2$.

• A. $\frac{1200}{343}$
• B. $\frac{3611}{489}$
• C. $\frac{3600}{361}$
• D. $\frac{4410}{349}$

Answer 1

The correct option is C $\frac{3600}{361}$

$p+q$ = $\frac{7-\sqrt{11}}{7+\sqrt{11}}$ + $\frac{7+\sqrt{11}}{7-\sqrt{11}}$
Rationalising the denominator of $p$ =$\frac{7-\sqrt{11}}{7+\sqrt{11}}$, we get,

$\frac{7-\sqrt{11}}{7+\sqrt{11}}\times \frac{7-\sqrt{11}}{7-\sqrt{11}}$ = $\frac{(7-\sqrt{11}{)}^2}{(7{)}^2-(\sqrt{11}{)}^2}$ = $\frac{(7-\sqrt{11}{)}^2}{49-11}$ = $\frac{(7-\sqrt{11}{)}^2}{38}$

Rationalising the denominator of $q$ =$\frac{7+\sqrt{11}}{7-\sqrt{11}}$, we get,
$\frac{7+\sqrt{11}}{7-\sqrt{11}}\times \frac{7+\sqrt{11}}{7+\sqrt{11}}$ = $\frac{(7+\sqrt{11}{)}^2}{(7{)}^2-(\sqrt{11}{)}^2}$ = $\frac{(7+\sqrt{11}{)}^2}{49-11}$ = $\frac{(7+\sqrt{11}{)}^2}{38}$

Now, $(p+q{)}^2$= $[\frac{(7-\sqrt{11}{)}^2}{38}+\frac{(7+\sqrt{11}{)}^2}{38}{]}^2$

= $\frac{1}{{38}^2}$ [$(7-\sqrt{11}{)}^2+(7+\sqrt{11}{)}^2{]}^2$

= $\frac{1}{{38}^2}$ [$49+11-14\sqrt{11}+49+11+14\sqrt{11}{]}^2$

=$\frac{(120{)}^2}{{38}^2}$

=$\frac{3600}{361}$