Answer: (1)

Solution

$ax^2 - 2bx + 5 = 0$

If $\alpha$ and $\alpha$ are roots of equations, then sum of roots: $2\alpha = \frac{2b}{a} \Rightarrow \alpha = \frac{b}{a}$

and product of roots $= \alpha^2 = \frac{5}{a}$, $\Rightarrow \frac{b^2}{a^2} = \frac{5}{a}$

$\Rightarrow b^2 = 5a$ …(i)

For $x^2 - 2bx - 10 = 0$: $\alpha + \beta = 2b$ …(ii) and $\alpha\beta = -10$ …(iii)

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

By eqn. (i) $\Rightarrow 5a - 10a^2 - 10a^2 = 0$ (substituting)

$\Rightarrow 20a^2 = 5a$

$\Rightarrow a = \frac{1}{4}$ and $b^2 = \frac{5}{4}$

$\alpha^2 = 20$ and $\beta^2 = 5$

Now, $\alpha^2 + \beta^2 = 5 + 20 = 25$