Question

Find the zeroes of the quadratic polynomial 9x${}^2$-6x-3 and verify the relationship between zeroes and the co-efficient of the polynomial.

Answer 1

Given quadratic polynomial is

$9x^2-6x-3=0\Rightarrow 9x^2-9x+3x-3=0\Rightarrow 9x(x-1)+3(x-1)=0\Rightarrow (9x+3)(x-1)=0\Rightarrow x=\frac{-1}{3}or1here,\alpha =\frac{-1}{3}and\beta =1also,a=9;b=-6;c=-3hence,\alpha +\beta =\frac{-1}{3}+1=\frac{2}{3}=\frac{-(-6)}{9}=\frac{-b}{a}and,\alpha .\beta =\frac{-1}{3}.1=\frac{-1}{3}=\frac{-3}{9}=\frac{c}{a}$

Hence proved