Question

If a, b, c are 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 $\gamma$ root that always satisfies:

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

Answer 1

Answer: (D)

$\alpha$ is a root of $a^2{x}^2+bx+c=0$
$\Rightarrow a^2{a}^2+ba+c=0$
$\Rightarrow b\alpha +c=-a^2{\alpha }^2...\left(i\right)$
$\beta$ is a root of $a^2{x}^2+bx+c=0$
$\Rightarrow b\beta +c=a^2{\beta }^2...\left(ii\right)$
Let $f\left(x\right)=a^2{x}^2+2bx+2c$. Then,
$f\left(\alpha \right)=a^2{\alpha }^2+2b\alpha +2c=a^2{\alpha }^2+2\left(b\alpha +c\right)$
$\Rightarrow f\left(\alpha \right)=a^2{\alpha }^2-2a^2{\alpha }^2$ [Using $\left(i\right)$]
$=-a^2{\alpha }^2<0$
and $f\left(\beta \right)=a^2{\beta }^2+2b\beta +2c$
$=a^2{\beta }^2+2\left(b\beta +c\right)$
$=a^2{\beta }^2+2a^2{\beta }^2=3a^2{\beta }^2>0$ [Using $\left(ii\right)$ ]
Thus, $f\left(x\right)$ is a polynomial such that f $\left(\alpha \right)<0$ and $f\left(\beta \right)>0$. Therefore, there exists $\gamma$ satisfying $\alpha <\gamma <\beta$ such that $f\left(\gamma \right)=0$