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 $\alpha^2+\beta^2=(\alpha+\beta)^2 -2\alpha\beta=4b^2+20=25.$
$b^2=5a$, 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 $\alpha^2+\beta^2=25.$
$$a\alpha^2-2b\alpha+5=0\tag{1}$$ $$\alpha^2-2b\alpha-10=0\tag{2}$$
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.
