Question

Let $a,b\in R$ be such that the equation $a{x}^2–2bx+15=0$ has a repeated root $\alpha$. If $\alpha$ and $\beta$ are the roots of the equation $x^2–2bx+21=0$, then ${\alpha }^2+{\beta }^2$ is equal to

• A. $37$
• B. $68$
• C. $58$
• D. $92$

Answer 1

The correct option is C $58$

$a{x}^2-2bx+15=0$ has repeated roots $\alpha$.
$\Rightarrow 2\alpha =\frac{2b}{a},{\alpha }^2=\frac{15}{a}$
$\Rightarrow \alpha \left(\frac{b}{a}\right)=\frac{15}{a}$
$\Rightarrow \alpha =\frac{15}{b}$

Now,
$x^2-2bx+21=0$
$\Rightarrow {\left(\frac{15}{b}\right)}^2-2b\left(\frac{15}{b}\right)+21=0$
$\Rightarrow {\left(\frac{15}{b}\right)}^2=9$
$\Rightarrow b^2=25$

$\alpha +\beta =2b,\alpha \beta =21$
${\alpha }^2+{\beta }^2=(\alpha +\beta {)}^2-2\alpha \beta$
$\Rightarrow {\alpha }^2+{\beta }^2=4b^2-42=58$