Question

Let$a,b$ and $c$ be real numbers $a\ne 0$, if $\alpha \ne 0.$ 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. $\gamma =a$
• B. $\alpha <\beta <\gamma$
• C. $\alpha <\gamma <\beta$
• D. $\gamma =\frac{[\alpha +\beta ]}{2}$

Answer 1

The correct option is C

$\alpha <\gamma <\beta$

The explanation for the correct option:

Finding the equation $a^2{x}^2+2bx-c=0$ has a root $\gamma$ that always satisfies:

Given $\alpha$ is a root of $a^2{x}^2+bx+c=0$, and $\beta$ is a root of $a^2{x}^2-bx-c=0$

Since $\alpha$ is a root of $a^2{x}^2+bx+c=0$

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

$\beta$ is a root of $a^2{x}^2–bx+c=0$

${\alpha }^2{\beta }^2–b\beta –c=0—\left(ii\right)$

Let $f\left(x\right)=a^2{x}^2+2bx+2c$

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

$=a^2{\alpha }^2–2a^2{\alpha }^2=–a^2{\alpha }^2$ [from equation (i)]

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

$=a^2{\beta }^2+2a^2{\beta }^2=3a^2{\beta }^2$ [from equation (ii)]

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

$f\left(x\right)$ must have a root lying in the open interval $(\alpha ,\beta ).$

$\alpha <\gamma <\beta$

Hence, the correct answer is option (C)