Question

Let $a,b,c$ be real numbers, $a\ne 0$. If $\alpha$ is a root of $a^2{x}^2+bx+c=0$, $\beta$ is a root of $a^2{x}^2-bx-c=0$ and $0<\alpha <\beta$, then the equation $a^2{x}^2+2bx+2c=0$ has a root $\gamma$ that always satisfies.

• A. $\gamma =\frac{[\alpha +\beta ]}{2}$
• B. $\gamma =\left[\alpha \right]+\left[\frac{\beta }{2}\right]$
• C. $\gamma =\left[\alpha \right]$
• D. $\alpha <\gamma <\beta$

Answer 1

The correct option is D

$\alpha <\gamma <\beta$

Explanation for the correct option:

Determining the correct option by finding the roots:

Given $f\left(x\right)=a^2{x}^2+2bx+2c=0....\left(i\right)$

Given that $\alpha$ is the root of $a^2{x}^2+bx+c=0$ then,

$$\begin{gathered} \Rightarrow a^2{\alpha }^2+b\alpha +c=0\dots \dots \left(ii\right) \\ \Rightarrow b\alpha +c=-a^2{\alpha }^2\dots ..\left(iii\right) \end{gathered}$$

And $\beta$ is a root of $a^2{x}^2-bx-c=0$ then

$\Rightarrow a^2{\beta }^2–b\beta –c=0...\left(iv\right)$

Substituting $x=\alpha$ in $\left(i\right)$

$f\left(\alpha \right)=a^2{\alpha }^2+2b\alpha +2c=0$

$$\begin{gathered} =(a^2{\alpha }^2+b\alpha +c)+(b\alpha +c) \\ =b\alpha +c\left[\because a^2{\alpha }^2+b\alpha +c=0\text{from equation}\left(ii\right)\right] \\ =–a^2{\alpha }^2\left[\text{from equation}\left(iii\right)\right] \end{gathered}$$

Substituting $x=\beta$ in $\left(i\right)$

$$\begin{gathered} f\left(\beta \right)=a^2{\beta }^2+2b\beta +2c \\ =3(b\beta +c) \\ =3{\alpha }^2{\beta }^2 \end{gathered}$$

Since $0<\alpha <\beta ,\alpha$ and $\beta$ are real.

$f\left(\alpha \right)<0,f\left(\beta \right)>0$

$f\left(\gamma \right)=0\text{where}\alpha <\gamma <\beta$

Hence, option (D) is the correct answer.