Question

Find the zeros of the quadratic polynomial $6x^2-3-7x$ and verify the relationship between the zeros and the coefficient of the polynomial.

Answer 1

$6x^2-3-7x=6x^2-7x-3=0$

$\Rightarrow 6x^2+2x-9x-3=0$

$\Rightarrow 2x(3x+1)-3(3x+1)=0$

$\Rightarrow (2x-3)(3x+1)=0$

Zeros = $\frac{3}{2},\frac{-1}{3}$

$\alpha +\beta =\frac{-b}{a}\Rightarrow \frac{3}{2}-\frac{1}{3}=\frac{7}{6}=\frac{(-7)}{6}=\frac{-b}{a}$

$\alpha \beta =\frac{c}{a}\Rightarrow \left(\frac{3}{2}\right)\left(\frac{-1}{3}\right)=\frac{-1}{2}=\frac{-3}{6}=\frac{c}{a}$

Hence proved.