Question

If $x=5-2\sqrt{6},$ find the value of $x^2+\frac{1}{x^2}$.

• A. $65$
• B. $98$
• C. $25$
• D. $78$

Answer 1

Answer: (B)

$98$

It is given that $x=5-2\sqrt{6}$.

Let us first find $x^2$ as follows:

$x^2={(5-2\sqrt{6})}^2={\left(5\right)}^2+{\left(2\sqrt{6}\right)}^2-(2\times 5\times \left(2\sqrt{6}\right))\{\because {(a-b)}^2=({a^2+b}^2-2ab)\}=25+(4\times 6)-20\sqrt{6}=25+24-20\sqrt{6}=49-20\sqrt{6}$

Therefore, $x^2=49-20\sqrt{6}$

Now consider $x^2+\frac{1}{x^2}$ as follows:

$x^2+\frac{1}{x^2}=49-20\sqrt{6}+\frac{1}{49-20\sqrt{6}}=\frac{{(49-20\sqrt{6})}^2+1}{49-20\sqrt{6}}=\frac{[{\left(49\right)}^2+{\left(20\sqrt{6}\right)}^2-(2\times 49\times \left(20\sqrt{6}\right))]+1}{49-20\sqrt{6}}\{\because {(a-b)}^2=({a^2+b}^2-2ab)\}=\frac{[2401+(400\times 6)-1960\sqrt{6}]+1}{49-20\sqrt{6}}$

$=\frac{2401+2400-1960\sqrt{6}+1}{49-20\sqrt{6}}=\frac{4802-1960\sqrt{6}}{49-20\sqrt{6}}=\frac{4802-1960\sqrt{6}}{49-20\sqrt{6}}\times \frac{49+20\sqrt{6}}{49+20\sqrt{6}}\left\{Rationalizedenominator\right\}=\frac{(4802-1960\sqrt{6})(49+20\sqrt{6})}{(49-20\sqrt{6})(49+20\sqrt{6})}$

$=\frac{(4802\times 49)+(4802\times 20\sqrt{6})+(-1960\sqrt{6}\times 49)+(-1960\sqrt{6}\times 20\sqrt{6})}{{\left(49\right)}^2-{\left(20\sqrt{6}\right)}^2}\{\because (a-b)(a+b)=a^2-b^2\}=\frac{235298+96040\sqrt{6}-96040\sqrt{6}-235200}{2401-2400}=\frac{98}{1}=98$

Hence, $x^2+\frac{1}{x^2}=98$ when $x=5-2\sqrt{6}$.