Question

Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients
$2\sqrt{3}{x}^2-5x+\sqrt{3}$

Answer 1

$$\begin{gathered} 2\sqrt{3}{x}^2-5x+\sqrt{3} \\ \Rightarrow 2\sqrt{3}{x}^2-2x-3x+\sqrt{3} \\ \Rightarrow 2x\left(\sqrt{3}x-1\right)-\sqrt{3}\left(\sqrt{3}x-1\right)=0 \end{gathered}$$

$$\begin{gathered} \Rightarrow \left(\sqrt{3}x-1\right)\mathrm{or}\left(2x-\sqrt{3}\right)=0 \\ \Rightarrow \left(\sqrt{3}x-1\right)=0\mathrm{or}\left(2x-\sqrt{3}\right)=0 \\ \Rightarrow x=\frac{1}{\sqrt{3}}\mathrm{or}x=\frac{\sqrt{3}}{2} \\ \Rightarrow x=\frac{1}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}=\frac{\sqrt{3}}{3}\mathrm{or}x=\frac{\sqrt{3}}{2} \end{gathered}$$

Sum of zeroes = $\frac{\sqrt{3}}{3}+\frac{\sqrt{3}}{2}=\frac{5\sqrt{3}}{6}=\frac{-\left(\mathrm{coefficient}\mathrm{of}x\right)}{\mathrm{coefficient}\mathrm{of}{x}^2}$ −4+(−3)=−71=−(coefficient of x)(coefficient of x2)
Product of zeroes = $\frac{\sqrt{3}}{3}\times \frac{\sqrt{3}}{2}=\frac{1}{2}=\frac{\mathrm{constant}\mathrm{term}}{\mathrm{coefficient}\mathrm{of}{x}^2}$ (−4)(−3)=121=constant termcoefficient of x2