Question

Let $a, b, c$ be real numbers, $a \neq 0$, if $\alpha$ is a root of $a^2 x^2+b x+c=0, \beta$ is a root of $a^2 x^2-b x-c=0$ and $0< \alpha< \beta$, then the equation $a^2 x^2+2 b x+2 c=0$ 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 $f(x)=a^2 x^2+2 b x+2 c$
$f (\alpha)= a ^2 \alpha^2+2 b \alpha+2 c = a ^2 \alpha^2-2 a ^2 \alpha^2=- a ^2 \alpha^2<0 $
$f (\beta)= a ^2 \beta^2+2 b \beta+2 c = a ^2 \beta^2+2 a ^2 \beta^2=3 a ^2 \beta^2>0$
Hence $f (\alpha)$ and $f (\beta)$ have opposite signs $\Rightarrow \alpha< \gamma< \beta$