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

$\therefore 2\alpha =\frac{2b}{a}\Rightarrow \alpha =\frac{b}{a}$and ${\alpha }^2=\frac{5}{a}\Rightarrow \frac{b^2}{a^2}=\frac{5}{a}$

$\Rightarrow b^2=5a ...\left(\mathrm{i}\right)$ $\left(a\ne 0\right)$

$\alpha +\beta =2b ...\left(\mathrm{ii}\right)$

and $\alpha \beta =-10 ...\left(\mathrm{iii}\right)$

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

$\Rightarrow b^2-2a{b}^2-10a^2=0$

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

$\Rightarrow 20a^2=5a$

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

Now ${\alpha }^2+{\beta }^2={\left(\alpha +\beta \right)}^2-2\alpha \beta$

$=5+20$

$=25$