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, $p(x^2-x)+x+5=0$
$\Rightarrow p{x}^2-(p-1)x+5=0$
$\Rightarrow \alpha +\beta =\frac{p-1}{p}$ and
$\alpha \beta =\frac{5}{p}$
Now, $\frac{\alpha }{\beta }+\frac{\beta }{\alpha }=\frac{4}{5}$
$\Rightarrow \frac{{\left(\alpha +\beta \right)}^2-2\alpha \beta }{\alpha \beta }=\frac{4}{5}$
$\Rightarrow \frac{{\left(p-1\right)}^2-10p}{5p}=\frac{4}{5}$
$\Rightarrow p^2-16p+1=0$
$\{\begin{array}{ll}& \text{p1+p2=16 }\\ & \text{ p1p2=1 }\end{array}$
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 $=254$