Question

Find the zeroes of the quadratic polynomial $9x^2-5$.

Answer 1

Given polynomial is $9x^2-5$

We know that, for a quadratic equation of the form $a{x}^2+bx+c=0$,

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Here, $a=9,b=0,c=-5$

$\therefore x=\frac{\pm \sqrt{180}}{18}$

$\Rightarrow x=\frac{\pm 6\sqrt{5}}{18}$

Hence, $x=\frac{6\sqrt{5}}{18}$ or $x=\frac{-6\sqrt{5}}{18}$

$\Rightarrow x=\frac{\sqrt{5}}{3}$ or $x=\frac{-\sqrt{5}}{3}$

Hence, $x=\frac{\sqrt{5}}{3},x=\frac{-\sqrt{5}}{3}$ is the solution of the given polynomial.