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?
 A: 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 $\alpha^2+\beta^2=(\alpha+\beta)^2 -2\alpha\beta=4b^2+20=25.$
A: $b^2=5a$, then $\alpha=b/a$
Next we have $\alpha+ \beta=2b$ and $\alpha \beta =-10$
Then $$\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha \beta=4b^2+20=20a+20 ~~~(*)$$
Next we have $$\beta=\frac{-10}{\alpha}=2b-\alpha \implies -\frac{10a}{b}=2b-\frac{b}{a}$$
$$\implies -10a^2=2ab^2-b^2 \implies -20a^2=-5a \implies a=\frac{1}{4},~ as ~a\ne 0$$ 
Finally from (*) we get $\alpha^2+\beta^2=25.$
A: $$a\alpha^2-2b\alpha+5=0\tag{1}$$
$$\alpha^2-2b\alpha-10=0\tag{2}$$
Subtracting $(1)$ and $(2)$
$$\alpha^2(a-1)+15=0$$
$$\alpha^2=\dfrac{15}{1-a}$$
From the first equation $$\alpha^2=\dfrac{5}{a}$$
$$\dfrac{15}{1-a}=\dfrac{5}{a}$$
$$3a=1-a$$
$$a=\dfrac{1}{4}$$
As first equation has equal roots
$$D=0$$
$$b^2-5a=0$$
$$b^2=\dfrac{5}{4}$$
So finally $\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta$
$$\alpha^2+\beta^2=(2b)^2+20=4b^2+20=25$$
So $25$ is your answer.
