Question

Sum of all real $x$ such that $\frac{4x^2+15x+17}{x^2+4x+12}=\frac{5x^2+16x+18}{2x^2+5x+13}$ is

• A. $-\frac{17}{4}$
• B. $-\frac{13}{4}$
• C. $-\frac{15}{3}$
• D. $-\frac{11}{3}$

Answer 1

The correct option is D $-\frac{11}{3}$

Let $f\left(x\right)=4x^2+15x+17,$ $g\left(x\right)=x^2+4x+12$ and $h\left(x\right)=x^2+x+1$
Then, the given equation becomes
$\frac{f\left(x\right)}{g\left(x\right)}=\frac{f\left(x\right)+h\left(x\right)}{g\left(x\right)+h\left(x\right)}$
$\Rightarrow f\left(x\right)g\left(x\right)+f\left(x\right)h\left(x\right)=f\left(x\right)g\left(x\right)+g\left(x\right)h\left(x\right)$
$\Rightarrow f\left(x\right)h\left(x\right)=g\left(x\right)h\left(x\right)$
Since $h\left(x\right)>0$ for all real $x,$ we may divide through by $h\left(x\right)$ to get
$f\left(x\right)=g\left(x\right)$
$\Rightarrow 4x^2+15x+17=x^2+4x+12$
$\Rightarrow 3x^2+11x+5=0$
The discriminant of this quadratic is ${11}^2-4\times 3\times 5=61>0$
So, it has two real roots
and sum of these roots is $-\frac{11}{3}$