Question

find the zeros and verify the relationship between the zeros and coefficients of the zeros of quadratic polynomial p(x)=x2-5 root 2x-12

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{p}\left(\mathrm{x}\right)={\mathrm{x}}^2-5\sqrt{2}\mathrm{x}-12 \\ \mathrm{Now},\mathrm{zeroes}\mathrm{of}\mathrm{p}\left(\mathrm{x}\right)\mathrm{are}\mathrm{given}\mathrm{by}, \\ \mathrm{p}\left(\mathrm{x}\right)=0 \\ \Rightarrow {\mathrm{x}}^2-5\sqrt{2}\mathrm{x}-12=0 \\ \Rightarrow {\mathrm{x}}^2-6\sqrt{2}\mathrm{x}+\sqrt{2}\mathrm{x}-12=0 \\ \Rightarrow \mathrm{x}\left(\mathrm{x}-6\sqrt{2}\right)+\sqrt{2}\left(\mathrm{x}-6\sqrt{2}\right)=0 \\ \Rightarrow \left(\mathrm{x}+\sqrt{2}\right)\left(\mathrm{x}-6\sqrt{2}\right)=0 \\ \Rightarrow \mathrm{x}+\sqrt{2}=0\mathrm{or}\mathrm{x}-6\sqrt{2}=0 \\ \Rightarrow \mathrm{x}=-\sqrt{2}\mathrm{or}\mathrm{x}=6\sqrt{2} \\ \mathrm{Now},\mathrm{sum}\mathrm{of}\mathrm{zeroes}=-\sqrt{2}+6\sqrt{2}=5\sqrt{2}=-\frac{\mathrm{coefficient}\mathrm{of}\mathrm{x}}{\mathrm{coefficient}\mathrm{of}{\mathrm{x}}^2} \\ \mathrm{product}\mathrm{of}\mathrm{zeroes}=-\sqrt{2}\times 6\sqrt{2}=-12=\frac{\mathrm{constant}\mathrm{term}}{\mathrm{coefficient}\mathrm{of}{\mathrm{x}}^2} \end{gathered}$$