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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?

prat
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1 Answers1

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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 $D$.

Matteo
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  • first of all, choice of $\alpha$ is questionable, because if we look at the other root $\dfrac{-b-\sqrt{b^2-4a^2c}}{2a^2}$, then this also suits as $\alpha$ if $b$ is negative and $c$ is positive.

    Another thing, root which you have choosen as $\alpha$ will be negative if $b$ and $c$ are positive, but it is given that $\alpha>0$

     – prat Mar 23 '20 at 16:02
  • @prat: yes, my error. I've corrected. Sorry. Also, in this case it isn't important the fact that $\alpha,\beta>0$. – Matteo Mar 23 '20 at 16:09
  • see as A,B,C options are getting ruled out, thats why we are saying answer is D), but what's the logic behind it. – prat Mar 23 '20 at 16:11
  • @prat: because, if you compute $\frac{\alpha+\beta}{2}$ or $\alpha+\frac{\beta}{2}$, you will never obtain $\gamma$ because you can't simplify the discriminant $b^2+4a^2c$ of $\beta$ that is not contained in $\gamma$. – Matteo Mar 23 '20 at 16:16
  • no I am not asking for verification of options $A,B,C$, I am asking why option D) is correct? – prat Mar 23 '20 at 16:18
  • @prat: I've shown you how to deduce that the answer is D not doing all things you have done, but using as you esclusion. – Matteo Mar 23 '20 at 16:21