Question

$f\left(x\right)=a{x}^3+b{x}^2+cx-3$ be a polynomial with real coefficient. Let $x_1,x_2,x_3,(x_i>0)$ be the roots of $f\left(x\right)=0$, such that $\sum _{i,j,k=1i\ne j\ne k}^3\frac{x_i}{x_j+x_k}=\frac{3}{2}$. Then which of following(s) is/are possible ?

• A. $a=\frac{1}{9}$
• B. $a=24$
• C. $f\left(x\right)=0$ has atleast one repeated root.
• D. $b>0$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $a=\frac{1}{9}$
B $a=24$
C $f\left(x\right)=0$ has atleast one repeated root.

$\sum _{i,j,k=1i\ne j\ne k}^3\frac{x_i}{x_j+x_k}=\frac{3}{2}$
$\Rightarrow \frac{x_1}{x_2+x_3}+\frac{x_2}{x_1+x_3}+\frac{x_3}{x_1+x_2}=\frac{3}{2}$
$\Rightarrow \frac{x_1}{x_2+x_3}+1+\frac{x_2}{x_1+x_3}+1+\frac{x_3}{x_1+x_2}+1=\frac{9}{2}$
$\Rightarrow (x_1+x_2+x_3)(\frac{1}{x_2+x_3}+\frac{1}{x_1+x_3}+\frac{1}{x_1+x_2})=\frac{9}{2}$
$.....\left(1\right)$

Now, consider three numbers $x_1+x_2,$ $x_2+x_3$ and $x_3+x_1$.
By using $\text{A.M.}\ge \text{H.M.}$
$\frac{(x_1+x_2)+(x_2+x_3)+(x_3+x_1)}{3}\ge \frac{3}{\frac{1}{x_2+x_3}+\frac{1}{x_1+x_3}+\frac{1}{x_1+x_2}}$
$\Rightarrow (x_1+x_2+x_3)(\frac{1}{x_2+x_3}+\frac{1}{x_1+x_3}+\frac{1}{x_1+x_2})\ge \frac{9}{2}$
$.....\left(2\right)$

From eqn $\left(1\right)\&\left(2\right),$
$\text{A.M.}=\text{H.M.}$
Hence, all the roots of $f\left(x\right)=0$ are equal.
$\therefore x_1=x_2=x_3=k$

Now, $f\left(x\right)=a{x}^3+b{x}^2+cx-3=0$
$\Rightarrow x_1+x_2+x_3=-\frac{b}{a}...\left(3\right)$
$\Rightarrow x_1{x}_2{x}_3=\frac{3}{a}...\left(4\right)$

When $k=\frac{1}{2}\Rightarrow a=24$
$k=3\Rightarrow a=\frac{1}{9}$

Since, $x_i>0\Rightarrow a>0$ [from eqn(4)]
Hence, from eqn(3), $b<0$