Question

Let $a, b\,\epsilon\,R, a \ne 0$ be such that the equation, $ax^{2}-2bx+5=0$ 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)

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