Question

The zeroes of a polynomial are -root 5 by 2 and -root 5 ; find the polynomial

Answer 1

Suppose $\alpha \mathrm{and}\beta$ are the roots of the polynomial P(x).
And the given roots are $\frac{-\sqrt{5}}{2}\mathrm{and}-\sqrt{5}$.
So suppose $\alpha =\frac{-\sqrt{5}}{2}\mathrm{and}\beta =-\sqrt{5}$
Then sum of the roots = $\alpha +\beta =\frac{-\sqrt{5}}{2}+\left(-\sqrt{5}\right)=\frac{-\sqrt{5}-2\sqrt{5}}{2}=\frac{-3\sqrt{5}}{2}...\left(\mathrm{i}\right)$
And product of the roots = $\alpha \beta =\frac{-\sqrt{5}}{2}\times \left(-\sqrt{5}\right)=\frac{5}{2}...\left(\mathrm{ii}\right)$
And we know that the equation having roots $\alpha \mathrm{and}\beta$ is given by;
$$\begin{gathered} x^2-\left(\alpha +\beta \right)x+\alpha \beta =0 \\ \Rightarrow x^2-\left(\frac{-3\sqrt{5}}{2}\right)x+\frac{5}{2}=0 \\ \Rightarrow x^2+\frac{3\sqrt{5}x}{2}+\frac{5}{2}=0\{\mathrm{using}(\mathrm{i})\mathrm{and}(\mathrm{ii}\left)\right\} \\ \Rightarrow 2x^2+3\sqrt{5}x+5=0 \end{gathered}$$
Therefore the required polynomial is $2x^2+3\sqrt{5}x+5=0$.