Question

Let $a,b\in R,a\ne 0,$ such that the equation, $a{x}^2-2bx+5=0$ has a repeated root $\alpha ,$ which is also a root of the equation $x^2-2bx-10=0.$ If $\beta$ is the other root of this equation, then ${\alpha }^2+{\beta }^2$ is equal to:

• A. $24$
• B. $25$
• C. $26$
• D. $28$

Answer 1

The correct option is B $25$

$a{x}^2-2bx+5=0$ has both roots as $\alpha$
$\Rightarrow 2\alpha =\frac{2b}{a}\Rightarrow \alpha =\frac{b}{a}$
And ${\alpha }^2=\frac{5}{a}$
$\Rightarrow b^2=5a(a\ne 0)\cdots \left(1\right)$
$\Rightarrow \alpha +\beta =2b$ and $\alpha \beta =-10$

$\beta ,\alpha$ are the roots of $x^2-2bx-10=0.$
$\alpha =\frac{b}{a}$
$\Rightarrow b^2-2a{b}^2-10a^2=0$
$\Rightarrow 5a-10a^2-10a^2=0\text{( Using (1)})$
$\Rightarrow a=\frac{1}{4}\Rightarrow b^2=\frac{5}{4}$
$\Rightarrow {\alpha }^2=20,{\beta }^2=5\Rightarrow {\alpha }^2+{\beta }^2=25$