Question

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

Answer 1

Let $f(x) = 4x^2 + 15x + 17,$ $g(x) = x^2 + 4x + 12$ and $h(x) = x^2 + x + 1$
Then, the given equation becomes
\(\dfrac {f(x)}{g(x)}=\dfrac{f(x)+h(x)}{g(x)+h(x)}\)
\(\Rightarrow f(x)g(x)+f(x)h(x) = f(x)g(x)+g(x)h(x)\)
\(\Rightarrow f(x)h(x)=g(x)h(x)\)
Since \(h(x) > 0\) for all real $x,$ we may divide through by $h(x)$ to get
\(f(x) = g(x)\)
\(\Rightarrow 4x^2+15x+17=x^2+4x+12\)
\(\Rightarrow 3x^2+11x+5=0\)
The discriminant of this quadratic is \(11^2-4\times3\times5=61>0\)
So, it has two real roots
and sum of these roots is $-\dfrac{11}{3}$