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 $f(\gamma )=0$