Question

If $x=9-4\sqrt{5},$ then find the value of $x^2-\frac{1}{x^2}$

Answer 1

Given $x=9-4\sqrt{5}$

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

We know that

$x-\frac{1}{x}=9-4\sqrt{5}-\frac{1}{9-4\sqrt{5}}=9-4\sqrt{5}-\left(\frac{9+4\sqrt{5}}{81-80}\right)$ [using rationlization]

$=9-4\sqrt{5}-9-4\sqrt{5}$

$=-8\sqrt{5}$

If $x-\frac{1}{x}=-8\sqrt{5}$ then

$x^2-\frac{1}{x^2}=-8\sqrt{5}(x+\frac{1}{x})$

Now,

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

$=18$.

So, $x^2-\frac{1}{x^2}=-8\sqrt{5}\left(18\right)=-144\sqrt{5}$

Hence $x^2-\frac{1}{x^2}=-144\sqrt{5}$.