Question

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

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

Answer 1

Answer: (D)

$\alpha <\beta <\gamma$

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

$\Rightarrow a^2{\alpha }^2+b\alpha +c=0$

$\Rightarrow b\alpha +c=-a^2{\alpha }^2$ .....(i)

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

$\Rightarrow b\beta +c=a^2{\beta }^2$ .....(ii)

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

Then, $a^2{\alpha }^2+2(b\alpha +c)=a^2{\alpha }^2-2a^2{\alpha }^2=-a^2{\alpha }^2<0$

and $f\left(\beta \right)=a^2{\beta }^2+2b\beta +2c>0$

$3a^2{\beta }^2>0$ (from Eq.(i))

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 $f\left(\gamma \right)=0$.