Let $a,b\epsilon R,a\ne0$ be such that the equation $ax^{2}-2bx+5=0$ 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 equations then $\alpha^{2}+\beta^{2}$ is equal to:

For ax^2 - 2bx + 5 = 0, repeated root alpha means discriminant = 0. (-2b)^2 - 4(a)(5) = 0 => 4b^2 = 20a => b^2 = 5a. Root alpha = -(-2b)/(2a) = b/a. Since alpha is a root of x^2 - 2bx - 10 = 0, (b/a)^2 - 2b(b/a) - 10 = 0. Substitute b^2 = 5a: (5a/a^2) - 2b^2/a - 10 = 0 => 5/a - 10 - 10 = 0 => 5/a = 20 => a = 1/4. Then b^2 = 5/4. Alpha = b/a = b/(1/4) = 4b. Alpha^2 = 16b^2 = 16(5/4) = 20. For the second equation, sum of roots alpha + beta = 2b, product alpha * beta = -10. Beta = -10/alpha. Alpha^2 + beta^2 = 20 + (-10/alpha)^2 = 20 + 100/alpha^2 = 20 + 100/20 = 20 + 5 = 25.