Question

Find the zeros of the quadratic polynomial (8 x² − 4) and verify the relation between the zeros and the coefficients.

Answer 1

$\begin{array}{l}\text{We have:}\\ f\left(x\right)=8x^2-4\\ \text{It can be written as 8}{x}^2+0x-4\\ =4\{(\sqrt{2}x{)}^2-(1{)}^2\}\\ =4(\sqrt{2}x+1)(\sqrt{2}x-1)\\ \therefore f\left(x\right)=0=>(\sqrt{2}x+1)(\sqrt{2}x-1)=0\\ =>\sqrt{2}x+1=0\text{or}\sqrt{2}x-1=0\\ =>x=\frac{-1}{\sqrt{2}}\text{or}x=\frac{1}{\sqrt{2}}\\ \text{So, the zeroes of}f\text{(}x\text{) are}\frac{-1}{\sqrt{2}}\text{and}\frac{1}{\sqrt{2}}\\ \text{Here the coefficient of}x\text{is 0 and the coefficient of}{x}^2\text{is}\sqrt{2}\\ \text{Sum of the zeroes}=\frac{-1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\frac{-1+1}{\sqrt{2}}=\frac{0}{\sqrt{2}}=\frac{-\left(\text{coefficient of}x\right)}{\left(\text{coefficient of}{x}^{\text{2}}\right)}\\ \text{Product of the zeroes=}\frac{-1}{\sqrt{2}}\times \frac{1}{\sqrt{2}}=\frac{-1\times 4}{2\times 4}=\frac{-4}{8}=\frac{\text{constant term}}{\text{(coefficient of}{x}^{\text{2}})}\end{array}$