Question

Let $a,bϵR,a\ne 0$ be such that the equation, $a{x}^2-2bx+5=0$ has a repeated root a, 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

Answer: (B)

$a{x}^2-2bx+5=0{<}_{\alpha }^{\alpha }$
$\Rightarrow \alpha =\frac{b}{a};{\alpha }^2=\frac{5}{a}\Rightarrow b^2=5a$
$x^2-2bx-10=0{<}_{\beta }^{\alpha }\Rightarrow {\alpha }^2-2b\alpha -10=0$
$\Rightarrow a=\frac{1}{4}\Rightarrow {\alpha }^2=20;\alpha \beta =-10\Rightarrow {\beta }^2=5$
$\Rightarrow {\alpha }^2+{\beta }^2=25$