Question

If $\alpha, \beta$ are roots of the equation $p (x^{2}-x)+x+5=0$ and $p_{1}, p_{2}$ are two values of p for which the roots $\alpha, \beta$ are connected by the relation $\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{4}{5}$, then the value of $\frac{p_{1}}{p_{2}}+\frac{p_{2}}{p_{1}}$ equals

• A. $254$
• B. $0$
• C. $245$
• D. $-254$

Answer 1

Answer: (A)

Given equation is, $p(x^{2}-x) +x+5=0$
$\Rightarrow px^{2}-(p-1)x+5=0$
$\Rightarrow \alpha+\beta=\frac{p-1}{p}$ and
$\alpha \beta=\frac{5}{p}$
Now, $\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{4}{5}$
$\Rightarrow \frac{\left(\alpha+\beta\right)^{2}-2\alpha\beta}{\alpha\beta}=\frac{4}{5}$
$\Rightarrow \frac{\left(p-1\right)^{2}-10p}{5p}=\frac{4}{5}$
$\Rightarrow p^{2}-16p+1=0$
$ \begin{cases} & \text {$p_{1}+p_{2}=16$ } \\[2ex] & \text { $p_{1}p_{2}=1$ } \end{cases}$
Now, $\frac{p_{1}}{p_{2}}+\frac{p_{2}}{p_{1}}$
$=\frac{\left(p_{1}+p_{2}\right)^{2}-2p_{1}p_{2}}{p_{1}p_{2}}$
$\therefore $ Required value is $= 254$