Question

If $\beta$ and $\frac{1}{\beta }$ are zeroes of the polynomial $({\alpha }^2+\alpha )x^2+61x+6\alpha$. Find the values of $\beta$ and $\alpha$.

Answer 1

Product of roots, $\beta \times \frac{1}{\beta }=\frac{6\alpha }{{\alpha }^2+\alpha }\Rightarrow 6\alpha ={\alpha }^2+\alpha$

$\Rightarrow {\alpha }^2-5\alpha =0$

$\Rightarrow \alpha (\alpha -5)=0\Rightarrow \alpha =0$ or $\alpha =5$

$(\beta =0)$

Sum of roots, $\beta +\frac{1}{\beta }=\frac{-61}{25+5}\Rightarrow ({\beta }^2+1)30=-61\beta$

$\Rightarrow 30\beta +61\beta +30=0$

$\Rightarrow \beta =\frac{-61\pm \sqrt{121}}{60}=\frac{-61\pm 11}{60}=\frac{-72}{60}$ or $\frac{-50}{60}$

$=-\frac{6}{5}$ or $-\frac{5}{6}$