Question

Find the zeros of the quadratic polynomial (6x² − 7x − 3) and verify the relation between its zeros an coefficients.

Answer 1

$\begin{array}{l}\text{We have:}\\ f\left(x\right)=6x^2-7x-3\\ =6x^2-9x+2x-3\\ =3x(2x-3)+1(2x-3)\\ =(3x+1)(2x-3)\\ \therefore f\left(x\right)=0=>(3x+1)(2x-3)=0\\ =>3x+1=0\text{or 2}x-3=0\\ =>x=\frac{-1}{3}\text{or}x=\frac{3}{2}\\ \text{So, the zeroes of}f\text{(}x\text{) are}\frac{-1}{3}\text{and}\frac{3}{2}\text{.}\\ \text{Sum of the zeroes}=\left(-\frac{1}{3}\right)+\frac{3}{2}=\frac{-2+9}{6}=\frac{7}{6}=\frac{-\left(\text{coefficient of}x\right)}{\left(\text{coefficient of}{x}^{\text{2}}\right)}\\ \text{Product of the zeros=}\left(-\frac{1}{3}\right)\times \frac{3}{2}\text{=}-\frac{3}{6}=\frac{\text{constant term}}{\text{(coefficient of}{x}^{\text{2}})}\end{array}$