Question

The number of real roots of the equation $\frac{4x}{x^2+x+3}+\frac{5x}{x^2-5x+3}=\frac{-3}{2}$ is equal to

• A. $0$
• B. $1$
• C. $2$
• D. $4$

Answer 1

Answer: (C)

$x=0$ is not a root.
Dividing both the numerators and denominators by $x$ and putting $y=x+\frac{3}{x}$ , we get,
$\frac{4}{y+1}+\frac{5}{y-5}=\frac{-3}{2}\Rightarrow y=-5,3$
$x+\frac{3}{x}=-5\Rightarrow x^2+5x+3=0\to$ two real roots
$x+\frac{3}{x}=3\Rightarrow x^2-3x+3=0\to$ two imaginary roots
$\Rightarrow$ Total two real roots