Question

$abc\ne 0$ & $a,b,c\in R$. If $x_1$ is a root of $a^2{x}^2+bx+c=0,x_2$ is a root of $a^2{x}^2-bx-c=0$ and $x_1>x_2>0$, then the equation $a^2{x}^2+2bx+2c=0$ has a root $x_3$ such that

• A. $x_3>x_1>x_2$
• B. $x_2>x_1>x_3$
• C. $x_1>x_3>x_2$
• D. $x_1>x_2>x_3$

Answer 1

The correct option is C

$x_1>x_3>x_2$

$x_1$ is a root of $a^2{x}^2+bx+c=0\Rightarrow a^2{x}_1^2+b{x}_1+c=0$
$x_2$ is a root of $a^2{x}^2-bx-c=0\Rightarrow a^2{x}_2^2-b{x}_2-c=0$
Let $f\left(x\right)=a^2{x}^2+2bx+2c$
Put $x=x_1$:
$f\left(x_1\right)=a^2{x}_1^2+2b{x}_1+2c$
$=-a^2{x}_1^2$
Now, put $x=x_2$
$f\left(x_2\right)=a^2{x}_2^2+2b{x}_2+2c$
$=3a^2{x}_2^2$
$f\left(x_1\right).f\left(x_2\right)=\left(3a^2{x}_2^2\right)(-a^2{x}_1^2)<0$
$\therefore$ One root of $a^2{x}^2+2bx+2c=0$ will lie between $x_1\&x_2$.
$\Rightarrow x_1>x_3>x_2(\because x_1>x_2)$