Question

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

• A. $\alpha =-5$ and $\beta =-6$
• B. $\alpha =5$ and $\beta \in R$
• C. $\alpha =5$ and $\beta =3$
• D. None of the above

Answer 1

The correct option is B $\alpha =5$ and $\beta \in R$

Solution:-

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

$\therefore$

$a=({\alpha }^2+\alpha )$

$b=61$

$c=6\alpha$

Product of roots $=\frac{c}{a}$

$\therefore \beta \times \frac{1}{\beta }=\frac{6\alpha }{{\alpha }^2+\alpha }$

$\Rightarrow 1=\frac{6\alpha }{{\alpha }^2+\alpha }$

$\Rightarrow {\alpha }^2+\alpha -6\alpha =0$

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

$\Rightarrow \alpha =0\text{or}5$

$\because a\ne 0$

$\Rightarrow {\alpha }^2+\alpha \ne 0$

$\Rightarrow \alpha \ne 0$

$\therefore \alpha =5$

Sum of roots $=\frac{-b}{a}$

$\therefore \beta +\frac{1}{\beta }=\frac{-61}{{\alpha }^2+\alpha }$

$\because \alpha =5$,

$\Rightarrow {\alpha }^2+\alpha =5^2+5=25+5=30$

$\therefore \beta +\frac{1}{\beta }=\frac{-61}{30}$

$\Rightarrow 30{\beta }^2+30=-61\beta$

$\Rightarrow 30{\beta }^2+61\beta +30=0$

From quadratic formula,

$\beta =\frac{-61\pm \sqrt{{61}^2-4\times 30\times 30}}{2\times 30}$

$\Rightarrow \beta =\frac{-61\pm 11}{60}$

$\Rightarrow \beta =\frac{-5}{6}\text{or}\frac{-6}{5}$

Since, value of $\beta$ does not match in any option.

The answer is none of these.