Let $a,b,c$ be real numbers, $a\ne0$. If a is a root of $a^2x^2++bx+c=0$, $\beta$ is the root of $a^2x^2-bx-c=0$ abd $0<\alpha<\beta$, then the equation $a^2x^2+2bx+2c=0$ has a root $\gamma$ that always satisfies.
A) $\gamma=\dfrac{\alpha+\beta}{2}$, B) $\gamma=\alpha+\dfrac{\beta}{2}$, C) $\gamma=\alpha$, D) $\alpha<\gamma<\beta$
My attempt is as follows:-
$$a^2\alpha^2+b\alpha+c=0\tag{1}$$ $$a^2\beta^2-b\beta-c=0\tag{2}$$
$$a^2(\alpha^2-\beta^2)+b(\alpha+\beta)+2c=0\tag{3}$$
Now checking every option
A) $\gamma=\dfrac{\alpha+\beta}{2}$
$$\dfrac{a^2(\alpha+\beta)^2}{4}+b(\alpha+\beta)+2c=0$$
This looks like equation $3$, but in equation $3$ coefficient of $a^2$ is negative and here it is positive, so first option is wrong.
B)$\gamma=\dfrac{2\alpha+\beta}{2}$
$$\dfrac{a^2(2\alpha+\beta)^2}{4}+b(2\alpha+\beta)+2c=0$$
This can look like equation $3$ but only for certain values of $\alpha$ and $\beta$, so this option is wrong.
C) $\gamma=\alpha$
$$a^2\alpha^2+2b\alpha+2c=0$$
This can look like equation $1$ but only for certain values of $\alpha$ and $\beta$, so this option is wrong.
So last option $D)$ will be the correct answer and it is indeed the correct option.
But I am not satisfied with my approach, so is there any approach which is better?
You can easily find the solutions to your equations (now I choose only the correct ones): $$\alpha=\frac{-b\pm\sqrt{b^2-4a^2c}}{2a^2}$$$$\beta=\frac{-b\pm\sqrt{b^2+4a^2c}}{2a^2}$$$$\gamma=\frac{-b\pm\sqrt{b^2-4a^2c}}{a^2}$$ Now, if you observe the $\Delta$, you can notice that are different, so possibilities $A,B$ are incorrect because if you sum in every possible way $\alpha$ and $\beta$ you will never obtain $\gamma$. But also $C$ is incorrect because $\gamma\neq\alpha$, so the right answer is $D$.