Question

$$\begin{gathered} \text{Q. Find the zero of the following polynomials by factorization method and verify the} \\ \text{relation between the zeros and the coefficients of the polynomial :} \\ ({\text{i) 2x}}^2\text{- ( 1 +}\sqrt[2]{2}\text{) x +}\sqrt{2} \end{gathered}$$

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{p}\left(\mathrm{x}\right)=2{\mathrm{x}}^2-\left(1+2\sqrt{2}\right)\mathrm{x}+\sqrt{2} \\ \mathrm{Now},\mathrm{zeroes}\mathrm{of}\mathrm{p}\left(\mathrm{x}\right)\mathrm{are}\mathrm{given}\mathrm{by}\mathrm{solving} \\ \mathrm{p}\left(\mathrm{x}\right)=0 \\ \Rightarrow 2{\mathrm{x}}^2-\left(1+2\sqrt{2}\right)\mathrm{x}+\sqrt{2}=0 \\ \Rightarrow 2{\mathrm{x}}^2-\mathrm{x}-2\sqrt{2}\mathrm{x}+\sqrt{2}=0 \\ \Rightarrow \mathrm{x}\left(2\mathrm{x}-1\right)-\sqrt{2}\left(2\mathrm{x}-1\right)=0 \\ \Rightarrow \left(\mathrm{x}-\sqrt{2}\right)\left(2\mathrm{x}-1\right)=0 \\ \Rightarrow \mathrm{x}-\sqrt{2}=0\mathrm{and}2\mathrm{x}-1=0 \\ \Rightarrow \mathrm{x}=\sqrt{2}\mathrm{and}\mathrm{x}=\frac{1}{2} \\ \mathrm{Now},\mathrm{sum}\mathrm{of}\mathrm{zeroes}=\sqrt{2}+\frac{1}{2}=\frac{2\sqrt{2}+1}{2}=-\frac{\mathrm{coefficient}\mathrm{of}\mathrm{x}}{\mathrm{coefficient}\mathrm{of}{\mathrm{x}}^2} \\ \mathrm{Product}\mathrm{of}\mathrm{zeroes}=\sqrt{2}\times \frac{1}{2}=\frac{\sqrt{2}}{2}=\frac{\mathrm{constant}\mathrm{term}}{\mathrm{coefficient}\mathrm{of}{\mathrm{x}}^2} \end{gathered}$$