If both roots of the equation $ax^2-2bx+5=0$ are $\alpha$ and roots of the equation $x^2-2bx-10=0$ are $\alpha$ and $\beta$. 
find $\alpha^2+\beta^2$

Both equations have a common root 
$$(-10a-5)^2=(-2ab+2b)(20b+10b)$$
$$25+100a^2+100a=60b^2(1-a)$$
Also since the first equation has  equal roots 
$$4b^2-20a=
0$$
$$b^2=5a$$
I could substitute the value of a in the above equation, but that gives me a biquadractic equation in b, and I don’t think it’s supposed to go that way. What am I doing wrong?

From the first equation, $\alpha^2=\frac{5}{a},2\alpha=\frac{2b}{a}$ and we obtain $\alpha =\frac{5}{b}$. 
From the second equation
$\alpha + \beta=2b, \alpha\beta=-10$. Then $\beta=-2b$ and $\frac{5}{b}=4b$ i.e. 
$b^2=\frac{5}{4}.$
Then $\alpha^2+\beta^2=(\alpha+\beta)^2 -2\alpha\beta=4b^2+20=25.$
