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