Question

Let $a, b, c$ be real number $\left(a \ne 0\right)$ 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 root of the equation (say $\gamma)$ $a^{2}\,x^{2}+2bx+2c=0$ 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)

Since $\alpha$ is a root of $a^{2}x^{2}+bx+c=0$
$\Rightarrow a^{2}\alpha^{2}+b \alpha+c=0$
$\therefore b \alpha+c=-a^{2} \alpha^{2}\, \dots (i)$
$\beta$ is a root of $a^{2}x^{2}-bx-c=0$
$\Rightarrow a^{2}\beta^{2}-b \beta-c=0$
$\Rightarrow a^{2}\beta^{2}=b \beta+c\, \dots(ii)$
Now, $f (x) =a^{2}x^{2}+2bx+2c$
$\therefore f (\alpha) = a^{2}\alpha^{2}+2(b \alpha+c)$
$=a^{2}\alpha^{2}+2(-\alpha^{2}-a^{2})=-a^{2}\alpha^{2}<\,0$ (From (i))
Similarly, $f (\beta) =3\alpha^{2}\beta^{2}>\,0$
Now $f (\alpha)$ and $f (\beta)$ arc of opposite sign and $0<\,\alpha<\,\beta$ (given)
$\therefore \exists$ exactly one real value between $\alpha$ and $\beta$ say $\gamma$ such that $f (\gamma) =0$