Question

Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients
$4x^2-4x+1$

Answer 1

$$\begin{gathered} 4x^2-4x+1=0 \\ \Rightarrow {\left(2x\right)}^2-2\left(2x\right)\left(1\right)+{\left(1\right)}^2=0 \\ \Rightarrow {\left(2x-1\right)}^2=0\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\left[\because a^2-2ab+b^2={\left(a-b\right)}^2\right] \end{gathered}$$

$$\begin{gathered} \Rightarrow {\left(2x-1\right)}^2=0\hspace{0.17em} \\ \Rightarrow x=\frac{1}{2}\mathrm{or}x=\frac{1}{2} \end{gathered}$$

Sum of zeroes = $\frac{1}{2}+\frac{1}{2}=1=\frac{1}{1}=\frac{-\left(\mathrm{coefficient}\mathrm{of}x\right)}{\mathrm{coefficient}\mathrm{of}{x}^2}$ −4+(−3)=−71=−(coefficient of x)(coefficient of x2)
Product of zeroes = $\frac{1}{2}\times \frac{1}{2}=\frac{1}{4}=\frac{\mathrm{constant}\mathrm{term}}{\mathrm{coefficient}\mathrm{of}{x}^2}$ (−4)(−3)=121=constant termcoefficient of x2