Let $a,b\in R,a\ne 0$  be such that the equation,  $a{x}^2-2bx+5=0$  has a repeated root  $\alpha ,$  which is also a root of the equation, $x^2-2bx-10=0.$   If $\beta$   is the other root of this equation, then $\left|\sqrt{{\alpha }^2+{\beta }^2}\right|$  is equal to

 $a{x}^2-2bx+5=0,$
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  $={\alpha }^2=\frac{5}{a}\Rightarrow \frac{b^2}{a^2}=\frac{5}{a}$
 $\Rightarrow b^2=5a(a\ne 0)$
For $x^2-2bx-10=0;\alpha +\beta =2b$   and  $\alpha \beta =-10$
and  $\alpha \beta =-10$
 $\alpha =\frac{b}{a}$ is also root of  $x^2-2bx-10=0$
 $\Rightarrow b^2-2a{b}^2-10a^2=0$
By eqn. (i)  $\Rightarrow 5a-10a^2-10a^2=0$
$\Rightarrow 20a^2=5a\Rightarrow a=\frac{1}{4}$  and  $b^2=\frac{5}{4}$
${\alpha }^2=20$  and  ${\beta }^2=5;$ Now,  ${\alpha }^2+{\beta }^2=5+20=25$