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?
Best Answer
$$a\alpha^2-2b\alpha+5=0\tag{1}$$ $$\alpha^2-2b\alpha-10=0\tag{2}$$
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.