Question

If $x=\frac{\sqrt{5}-2}{\sqrt{5}+2}$, $y=\frac{\sqrt{5}+2}{\sqrt{5}-2}$ and $y^2$ is $(161+72\sqrt{m})$, then value of $m$ is ___.

Answer 1

We have $x=\frac{\sqrt{5}-2}{\sqrt{5}+2}$ and $y=\frac{\sqrt{5}+2}{\sqrt{5}-2}$

Rationalizing the denominator gives,

$y=\frac{\sqrt{5}+2}{\sqrt{5}-2}\times \frac{\sqrt{5}+2}{\sqrt{5}+2}$

$y=\frac{5+4+4\sqrt{5}}{5-4}$ (By applying the identities)

$y=\frac{9+4\sqrt{5}}{1}$

$y^2=(9+4\sqrt{5})(9+4\sqrt{5})$

$=81+16\times 5+72\sqrt{5}$

$=161+72\sqrt{5}$

Comparing this with $y^2=161+72\sqrt{m}$, we get $m=5$.