$\text{ Let }a·b\in R,a\ne 0\text{ be such that the equation, }a{x}^2-2bx+5=0\text{ has a repeated root }\alpha \text{ , which is }$

$\text{ also a root of the equation, }{x}^2-2bx-10=0\text{ . If }\beta \text{ is the other root of this equation, then }$

${\alpha }^2+{\beta }^2\text{ is equal to: }$

$\begin{array}{l}a{x}^2-2bx+5=0\text{ roots }\alpha ,\alpha \\ 2\alpha =\frac{2b}{a},{\alpha }^2=\frac{5}{a} x^2-2bx-10=0\\ \alpha =\frac{b}{a},\frac{b^2}{a^2}=\frac{5}{a} \text{ roots }\alpha ,\beta \alpha +\beta =2b\\ b^2=5a, \alpha \beta =-10\\ \alpha \text{ is a root of }{x}^2-2bx-10=0\\ {\alpha }^2-2b\alpha -10=0\\ \frac{b^2}{a^2}-2b\frac{b}{a}-10=0\\ \frac{b^2}{a^2}-\frac{2b^2}{a}-10=0\\ \frac{5}{a}-10-10=0\Rightarrow a=\frac{1}{4}\\ b^2=5a\Rightarrow b^2=\frac{5}{4}\\ {\alpha }^2+{\beta }^2=(\alpha +\beta {)}^2-2\alpha \beta =4b^2+20\\ 4·\frac{5}{4}+20=25\end{array}$