Question

The sum of all real value of $x$ for which $\frac{3x^2-9x+17}{x^2+3x+10}=\frac {5x^2-7x+19}{3x^2+5x+12}$ is equal to________.

Answer 1

Given,

$\frac{3x^2-9x+17}{x^2+3x+10}=\frac {5x^2-7x+19}{3x^2+5x+12}$

⇒ $\frac{x^2+3x+10+2x^2-12x+7}{x^2+3x+10}=\frac {3x^2+5x+12+2x^2-12x+7}{3x^2+5x+12}$

⇒ $1+\frac {2x^2-12x+7}{x^2+3x+10}=1+\frac {2x^2-12x+7}{3x^2+5x+12}$

⇒ $2x^2-12x+7 \frac{1}{x^2+3x+10}-\frac {1}{3x^2+5x+12}=0$

$2x^2-12x+7=0$ Or $3x^2+5x+12=x^2+3x+10$

For $2x^2-12x+7=0$

Sum of roots=$\frac {12}{2}=6$

Or $3x^2+5x+12=x^2+3x+10$

⇒ $2x^2+2x+2=0$

⇒ $x^2+x+1=0$

No real roots as $D=1-4<0$