Question

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

• A. $\gamma =\alpha$
• B. $\gamma =\beta$
• C. $\gamma =(\alpha +\beta )/2$
• D. $\alpha <\gamma <\beta$

Answer 1

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

Let $f\left(x\right)=a^2{x}^2+2bx+2c$
From the question, $a^2{\alpha }^2+b\alpha +c=0and{a}^2{\beta }^2-b\beta -c=0$
Now, $f\left(\alpha \right)=a^2{\alpha }^2+2b\alpha +2c=b\alpha +c=-a^2{\alpha }^{2}f\left(\beta \right)=a^2{\beta }^2+2b\beta +2c=3(b\beta +c)=3a^2{\beta }^2$
but $0<\alpha <\beta \Rightarrow \alpha ,\beta$ are real
$\therefore f\left(\alpha \right)<0,f\left(\beta \right)>0$
Hence, $\alpha <\gamma <\beta .$