Let $a, b \in R, a \neq 0$ be such that the equation , $a x^{2} - 2 b x + 1 = 0$ has a repeated root $α$ , which is also a root of the equation, $x^{2} - 2 b x - 1 = 0$ . If $β$ is the other root of this equation , then $α^{2} + β^{2}$ is equal to

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$\frac{17}{4}$

$\frac{5}{2}$

$\frac{10}{3}$

answer is B.

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Detailed Solution

Discriminant = 0 $\Rightarrow 4 b^{2} - 4 a = 0 \Rightarrow b^{2} = a ..... (1)$
also , $α = \frac{2 b}{2 a} = \frac{1}{b}$
$α$ satisfies 2nd equation
$∴ \frac{1}{b^{2}} - 2 - 1 = 0 \Rightarrow b = \pm \sqrt{3} = \frac{1}{α}$
$α β = - 1 \Rightarrow β = \frac{- 1}{α}$
$∴ α^{2} + β^{2} = \frac{1}{3} + 3 = \frac{10}{3}$