Let a, b ∈ R, a ≠ 0 be such that the equation, a x ^ 2 - 2bx + 5 =...

Given, $a{x}^{2} - 2bx + 5 = 0$ has repeated root $\alpha$.

$∴ 2\alpha = \dfrac{2b}{a} ⇒ \alpha = \dfrac{b}{a}$and ${\alpha}^{2} = \dfrac{5}{a} ⇒ \dfrac{{b}^{2}}{{a}^{2}} = \dfrac{5}{a}$

$⇒ {b}^{2} = 5a . . . \left(i\right) \left(a \ne 0\right)</p><p>$\alpha + \beta = 2b . . . \left(ii\right)$</p><p>and$\alpha\beta = - 10 . . . \left(iii\right)

$\alpha =$\dfrac{b}{a} is also root$of {x}^{2} - 2bx - 10 = 0$

$⇒ {b}^{2} - 2a{b}^{2} - 10{a}^{2} = 0$

by \left(i\right)$⇒ 5a - 10{a}^{2} - 10{a}^{2} = 0$

$⇒ 20{a}^{2} = 5a$

$⇒ a =$\dfrac{1}{4}$and {b}^{2}$= \dfrac{5}{4}

Now ${\alpha}^{2} + {\beta}^{2} = {\left(\alpha + \beta\right)}$^{2} - 2\alpha\beta$</p><p>$= 5 + 20$</p><p>$= 25$</p>$