Theory of Equations 5 Question 5
5. 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 the 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
$(1989,2 M)$
(a) $\gamma=\frac{\alpha+\beta}{2}$
(b) $\gamma=\alpha+\frac{\beta}{2}$
(c) $\gamma=\alpha$
(d) $\alpha<\gamma<\beta$
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Answer:
Correct Answer: 5. (d)
Solution:
- Since, $\alpha$ is a root of $a^{2} x^{2}+b x+c=0$
$\Rightarrow \quad a^{2} \alpha^{2}+b \alpha+c=0$
and $\beta$ is a root of $\quad a^{2} x^{2}-b x-c=0$
$\Rightarrow \quad a^{2} \beta^{2}-b \beta-c=0$
Let $\quad f(x)=a^{2} x^{2}+2 b x+2 c$
$\therefore \quad f(\alpha)=a^{2} \alpha^{2}+2 b \alpha+2 c$
$ =a^{2} \alpha^{2}-2 a^{2} \alpha^{2}=-a^{2} \alpha^{2} $
[from Eq. (i)]
and $\quad f(\beta)=\alpha^{2} \beta^{2}+2 b \beta+2 c$
$ =a^{2} \beta^{2}+2 a^{2} \beta^{2}=3 a^{2} \beta^{2} \text { [from Eq. (ii)] } $
$ \Rightarrow \quad f(\alpha) f(\beta)<0 $
$f(x)$ must have a root lying in the open interval $(\alpha, \beta)$.
$ \therefore \quad \alpha<\gamma<\beta $
