Question

Find the zeros of the polynomial 9x² − 5 and verify the relation between zeros and coefficients.

Answer 1

$f\left(x\right)=9x^2-5={\left(3x\right)}^2-{\left(\sqrt{5}\right)}^2=\left(3x+\sqrt{5}\right)\left(3x-\sqrt{5}\right)$
∴ $f\left(x\right)=0\Rightarrow \left(3x+\sqrt{5}\right)\left(3x-\sqrt{5}\right)=0$
$\Rightarrow \left(3x+\sqrt{5}\right)=0\mathrm{or}\left(3\mathrm{x}-\sqrt{5}\right)=0$
$\Rightarrow \mathrm{x}=\frac{-\sqrt{5}}{3}\mathrm{or}\mathrm{x}=\frac{\sqrt{5}}{3}$
So, the zeros of f(x) are $-\frac{\sqrt{5}}{3}\mathrm{and}\frac{\sqrt{5}}{3}$.
$\mathrm{Sum}\mathrm{of}\mathrm{zeros}=\left(\frac{-\sqrt{5}}{3}+\frac{\sqrt{5}}{3}\right)=0=\frac{0}{9}=\frac{-\left(\mathrm{coefficient}\mathrm{of}\mathrm{x}\right)}{\left(\mathrm{coefficient}\mathrm{of}{\mathrm{x}}^2\right)}$
$\mathrm{Product}\mathrm{of}\mathrm{zeros}=\left(\frac{-\sqrt{5}}{3}\times \frac{\sqrt{5}}{3}\right)=\frac{-5}{9}=\frac{\left(\mathrm{constant}\mathrm{term}\right)}{\left(\mathrm{coefficient}\mathrm{of}{\mathrm{x}}^2\right)}$