Question

Find the zeros of the quadratic polynomial (x² − 5) and verify the relation between the zeros and the coefficients.

Answer 1

$\begin{array}{l}\text{We have:}\\ f\left(x\right)=x^2-5\\ \text{It can be written as}{x}^2+0x-5.\\ =(x^2-(\sqrt{5}{)}^2)\\ =(x+\sqrt{5})(x-\sqrt{5})\\ \therefore f\left(x\right)=0=>(x+\sqrt{5})(x-\sqrt{5})=0\\ =>x+\sqrt{5}=0\text{or}x-\sqrt{5}=0\\ =>x=-\sqrt{5}\text{or}x=\sqrt{5}\\ \text{So, the zeroes of}f\text{(}x\text{) are}-\sqrt{5}\text{and}\sqrt{5}\text{.}\\ \text{Here, the coefficient of}x\text{is 0 and the coefficient of}{x}^2\text{is 1}\text{.}\\ \text{Sum of the zeroes}=-\sqrt{5}+\sqrt{5}=\frac{0}{1}=\frac{-\left(\text{coefficient of}x\right)}{\left(\text{coefficient of}{x}^{\text{2}}\right)}\\ \text{Product of the zeroes=}-\sqrt{5}\times \sqrt{5}=\frac{-5}{1}=\frac{\text{constant term}}{\text{(coefficient of}{x}^{\text{2}})}\end{array}$