Question

. Find the zeros of the polynomial sqrt5 x2 7x - 6sqrt5and verify the relationship between

the zeros and the coefficients of the polynomial

Answer 1

Considering the given quadratic polynomial as $\sqrt{5}{x}^2+7x-6\sqrt{5}$.
To find the zeroes of the polynomial equating it to 0 as;
$$\begin{gathered} \sqrt{5}{x}^2+7x-6\sqrt{5}=0 \\ \Rightarrow \sqrt{5}{x}^2+10x-3x-6\sqrt{5}=0 \\ \Rightarrow \sqrt{5}x\left(x+2\sqrt{5}\right)-3\left(x+2\sqrt{5}\right)=0 \\ \Rightarrow \left(\sqrt{5}x-3\right)\left(x+2\sqrt{5}\right)=0 \\ \Rightarrow \left(\sqrt{5}x-3\right)=0\mathrm{and}\left(x+2\sqrt{5}\right)=0 \\ \Rightarrow x=\frac{3}{\sqrt{5}}\mathrm{and}x=-2\sqrt{5} \end{gathered}$$
Therefore the zeroes of the given polynomial are $\frac{3}{\sqrt{5}}\mathrm{and}-2\sqrt{5}$.
Sum of the zeroes of the polynomial = $\frac{3}{\sqrt{5}}-2\sqrt{5}=\frac{-7}{\sqrt{5}}=\frac{-\left(\mathrm{coeff}.\mathrm{of}x\right)}{\mathrm{coeff}.\mathrm{of}{x}^2}$
And product of the zeroes = $\frac{3}{\sqrt{5}}\times \left(-2\sqrt{5}\right)=-6=\frac{\mathrm{const}.\mathrm{term}}{\mathrm{coeff}.\mathrm{o}f{x}^2}$