<s>[INST] 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?
 [/INST] $b^2=5a$, 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 $\alpha^2+\beta^2=25.$
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