Question

Find the zeroes of the quadratic polynomial $6x^2-13x+6$ and verify the relation between the zeroes and its coefficients?

Answer 1

Given, the polynomial

${6x}^2-13x+6=0\Rightarrow {6x}^2-9x-4x+6=0\Rightarrow 3x(2x-3)-2(3x-3)=0\Rightarrow (2x-3)(3x-2)=0\therefore 2x-3=0\Rightarrow x=\frac{3}{2}3x-2=0\Rightarrow x=\frac{2}{3}$

Let, $\alpha =\frac{3}{2},\beta =\frac{2}{3}$

From the polynomial we get, $a=6,b=-13,c=6$

$\therefore$ Relation between the zeroes and coefficients,

Sum of the zeroes=$\frac{3}{2}+\frac{2}{3}=\frac{13}{6}=-\frac{(-13)}{6}=\frac{b}{a}$

and product of the zeroes=$\frac{3}{2}\times \frac{2}{3}=\frac{6}{6}=\frac{c}{a}$

Hence, the relation between the zeroes and the coefficients of the polynomial is verified.