Question

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

• A. $\gamma =\frac{\alpha +\beta }{2}$
• B. $\gamma =\frac{\alpha -\beta }{2}$
• C. $\gamma =\frac{\alpha \beta }{2}$
• D. $\alpha <\gamma <\beta$

Answer 1

The correct option is D $\alpha <\gamma <\beta$

$\alpha$ is a root of $a^2{x}^2+bx+c=0$
$\Rightarrow a^2{\alpha }^2+b\alpha +c=0\Rightarrow b\alpha +c=-a^2{\alpha }^2...............\left(1\right)$
Similarly,
$\Rightarrow b\beta +c=a^2{\beta }^2...............\left(2\right)$
Let $f\left(x\right)=a^2{x}^2+2bx+2c$
Then,
$f\left(\alpha \right)=a^2{\alpha }^2+2b\alpha +2c=-a^2{\alpha }^2<0......\text{from}\left(1\right)f\left(\beta \right)=a^2{\beta }^2+2b\beta +2c=3a^2{\beta }^2>0......\text{from}\left(2\right)$
Thus, $f\left(x\right)$ is a polynomial function such that $f\left(\alpha \right)<0\&f\left(\beta \right)>0$
Therefore, there exists $\gamma$ satifying $\alpha <\gamma <\beta$ such that $f\left(\gamma \right)=0$