Question

For the equation $\frac{4x^2+x+4}{x^2+1}+\frac{x^2+1}{x^2+x+1}=\frac{31}{6}$
Which of the following statement(s) is/are correct?

• A. The equation has $4$ real and distinct roots.
• B. The equation has $3$ real and distinct roots.
• C. The sum of all real and distinct roots is $-2.$
• D. The sum of all real and distinct roots is $-1.$

Answer 1

Answer: (B), (C)

The correct options are
B The equation has $3$ real and distinct roots.
C The sum of all real and distinct roots is $-2.$

$\frac{4x^2+x+4}{x^2+1}+\frac{x^2+1}{x^2+x+1}=\frac{31}{6}\Rightarrow 4+\frac{x}{x^2+1}+\frac{1}{1+\frac{x}{x^2+1}}=\frac{31}{6}$
Assuming $\frac{x}{x^2+1}=t$
$\Rightarrow 4+t+\frac{1}{1+t}=\frac{31}{6}\Rightarrow \frac{(t+4)(t+1)+1}{(t+1)}=\frac{31}{6}\Rightarrow 6(t^2+5t+5)=31(t+1)\Rightarrow 6t^2-t-1=0\Rightarrow (3t+1)(2t-1)=0\Rightarrow t=-\frac{1}{3},\frac{1}{2}$

When $t=-\frac{1}{3}$
$\frac{x}{x^2+1}=-\frac{1}{3}\Rightarrow x^2+3x+1=0D=9-4=5>0$
Two real and distinct roots.

When $t=\frac{1}{2}$
$\frac{x}{x^2+1}=\frac{1}{2}\Rightarrow x^2-2x+1=0\Rightarrow (x-1{)}^2=0$
Two real and equal roots.

Hence, the given equation has $3$ real and distinct roots, and
The sum of distinct roots
$=-3+1=-2$