Question

If $x=\sqrt{6}+\sqrt{5},$ then $x^2+\frac{1}{x^2}-2=$

• A. $2\sqrt{6}$
• B. $2\sqrt{5}$
• C. 24
• D. 20

Answer 1

The correct option is D ($20$)

Given $x=\sqrt{6}+\sqrt{5}$ ......(i)
Applying reciprocal property,
Then $\frac{1}{x}=\frac{1}{\sqrt{6}+\sqrt{5}}=\frac{\sqrt{6}-\sqrt{5}}{(\sqrt{6}+\sqrt{5})(\sqrt{6}-\sqrt{5})}$

$=\frac{\sqrt{6}-\sqrt{5}}{(\sqrt{6}{)}^2-(\sqrt{5}{)}^2}=\frac{\sqrt{6}-\sqrt{5}}{6-5}$

$=\frac{\sqrt{6}-\sqrt{5}}{1}=$
$\therefore \frac{1}{x}=\sqrt{6}-\sqrt{5}$ ........(ii)

Now, consider $x^2+\frac{1}{x^2}-2={(x-\frac{1}{x})}^2$
[Since, $(a-b{)}^2=a^2+b^2-2ab$]

$=\left[\right(\sqrt{6}+\sqrt{5})-(\sqrt{6}-\sqrt{5}){]}^2$
[From equations (i) and (ii)]
$=(\sqrt{6}+\sqrt{5}-\sqrt{6}+\sqrt{5}{)}^2=(2\sqrt{5}{)}^2$
$=4\times 5=20$
Hence, the value of the given expression is $20$.