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

Answer 1

Answer: (D)

Given $a^2{\alpha }^2+b\alpha +c=0$ and $a^2{\beta }^2-b\beta -c=0$
Let $f(x)=a^2{x}^2+2bx+2c$
$f(\alpha )=a^2{\alpha }^2+2b\alpha +2c=a^2{\alpha }^2-2a^2{\alpha }^2=-a^2{\alpha }^2<0$
$f(\beta )=a^2{\beta }^2+2b\beta +2c=a^2{\beta }^2+2a^2{\beta }^2=3a^2{\beta }^2>0$
Hence $f(\alpha )$ and $f(\beta )$ have opposite signs $\Rightarrow \alpha <\gamma <\beta$