Question

If $p=5+2\sqrt{6}$ and $q=\frac{1}{p}$, find $p^2+q^2$ is :

• A. $49$
• B. $98$
• C. $100$
• D. $Noneofthese$

Answer 1

Answer: (B)

$98$

$p=5+2\sqrt{6}$

$\Rightarrow q=\frac{1}{p}=\frac{1}{5+2\sqrt{6}}$

Rationalizing $q$ by multiplying $5-2\sqrt{6}$ in both numerator and denominator,

$\Rightarrow q=\frac{1}{5+2\sqrt{6}}\frac{\left(5-2\sqrt{6}\right)}{\left(5-2\sqrt{6}\right)}=\frac{\left(5-2\sqrt{6}\right)}{25-4\left(\sqrt{6}\times \sqrt{6}\right)}=\frac{5-2\sqrt{6}}{25-24}=5-2\sqrt{6}$

$\therefore p^2+q^2={\left(5+2\sqrt{6}\right)}^2+{\left(5-2\sqrt{6}\right)}^2$

$=25+{\left(2\sqrt{6}\right)}^2+2\left(5\right)\left(2\sqrt{6}\right)+25+{\left(2\sqrt{6}\right)}^2-2\left(5\right)\left(2\sqrt{6}\right)$

$=50+48$

$=98$

Hence, the answer is $98.$