Question

Find the zeros of the quadratic polynomial 6x² − 3 − 7x and verify the relationship between the zeros and the coefficients of the given polynomial.

Answer 1

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