Question

Find the zeroes of the polynomial $x^2-3$ and verify the relationship between the zeroes and the coefficients.

Answer 1

We have,

$x^2-3=x^2-(\sqrt{3}{)}^2$

$=(x+\sqrt{3})(x-\sqrt{3})$

$\therefore x+\sqrt{3}=0$

$x=-\sqrt{3}$

$\therefore x-\sqrt{3}=0$

$x=\sqrt{3}$

Sum of Zeroes =$\left(\sqrt{3}\right)+(-\sqrt{3})$

$=\sqrt{3}-\sqrt{3}$

$=0$

Product of zeroes $=\left(\sqrt{3}\right)\times (-\sqrt{3})$

$=-{\sqrt{3}}^2$

$=-3$

$\therefore \frac{0}{1}=\frac{-\left(coefficientofx\right)}{coefficientof{x}^2}$

$-\frac{3}{1}=\frac{constantterm}{coefficientsof{x}^2}$

Hence, this is the answer.