Question

Show that $\sin {36}^{\circ }$ is a root of the equation $16x^4-20x^2+5=0$

Answer 1

Given equation:

$16x^4-20x^2+5=0$

Let $x^2=t$ then

$16t^2-20t+5=0$

Using quadratic formula, we get

$t=\frac{20\pm \sqrt{{20}^2-4\times 16\times 5}}{2\times 16}$

$t=\frac{20\pm \sqrt{400-320}}{32}$

$t=\frac{20\pm \sqrt{80}}{32}$

$t=\frac{4(5\pm \sqrt{5})}{32}$

$x^2=\frac{(5\pm \sqrt{5})}{8}$

$x^2=\sqrt{\frac{(5\pm \sqrt{5})}{8}}$

$x=\frac{\sqrt{10\pm 2\sqrt{5}}}{4}$

$\sin {36}^{\circ }=\frac{\sqrt{10-2\sqrt{5}}}{4}$

$\therefore \sin {36}^{\circ }$ is a root