Question

Given $x=9-4\sqrt{5}$, Evaluate $x^2+\frac{1}{x^2}$

• A. $322$
• B. $\sqrt{77}$
• C. $161$
• D. None of these

Answer 1

The correct option is A $322$

$x=9-4\sqrt{5}\frac{1}{x}=\frac{1}{9-4\sqrt{5}}\times \frac{9+4\sqrt{5}}{9+4\sqrt{5}}\text{( By rationalizing)}\frac{1}{x}=\frac{9+4\sqrt{5}}{(9{)}^2-(4\sqrt{5}{)}^2}=\frac{9+4\sqrt{5}}{81-80}=9+4\sqrt{5}\frac{1}{x}=9+4\sqrt{5}$

Now,

${(x+\frac{1}{x})}^2=x^2+\frac{1}{x^2}+2.x.\frac{1}{x}$

${(x+\frac{1}{x})}^2=x^2+\frac{1}{x^2}+2$

$x^2+\frac{1}{x^2}={(x+\frac{1}{x})}^2-2$

$x^2+\frac{1}{x^2}={\left(\right(9-4\sqrt{5})+(9+4\sqrt{5}\left)\right)}^2-2$

$x^2+\frac{1}{x^2}={(9+9)}^2-2$

$x^2+\frac{1}{x^2}={\left(18\right)}^2-2$

$x^2+\frac{1}{x^2}=324-2=322$

Option A is the correct answer.