Question

Let $\alpha ,\beta \in R$. If $\alpha ,{\beta }^2$ be the roots of quadratic equation $x^2-px+1=0$ and ${\alpha }^2,\beta$ be the roots of quadratic equation $x^2-qx+8=0$, then the value of ' $r$ ' if $\frac{r}{8}$ be arithmetic mean of $p$ and $q$, is

• A. $\frac{83}{8}$
• B. $\frac{83}{4}$
• C. $\frac{83}{2}$
• D. 83

Answer 1

Answer: (D)

For the equation $x^2-px+1=0$,
the product of roots, $\alpha {\beta }^2=1$
and for the equation $x^2-qx+8=0$,
the product of roots ${\alpha }^2\beta =8$
Hence, $\left(\alpha {\beta }^2\right)\left({\alpha }^2\beta \right)=8\Rightarrow {\alpha }^3{\beta }^3=8\Rightarrow \alpha \beta =2$
$\therefore$ From $\alpha {\beta }^2=1$, we have $\beta =\frac{1}{2}$ and from ${\alpha }^2\cdot \beta =8$, we have $\alpha =4$
Hence, from sum of roots $=-\frac{b}{a}$ " relation, we have $p=\alpha +{\beta }^2=4+\frac{1}{4}=\frac{17}{4}$
and $q={\alpha }^2+\beta =16+\frac{1}{2}=\frac{33}{2}$
$\because \frac{r}{8}$ is arithmetic mean of $p$ and $q$
$\therefore \frac{r}{8}=\frac{p+q}{2}\Rightarrow r=4(p+q)=4\left(\frac{17}{4}+\frac{33}{2}\right)=17+66=83$