JEE MAIN - Mathematics (2022 - 28th July Morning Shift - No. 24)
The sum of all real values of $$x$$ for which $$\frac{3 x^{2}-9 x+17}{x^{2}+3 x+10}=\frac{5 x^{2}-7 x+19}{3 x^{2}+5 x+12}$$ is equal to __________.
Answer
6
Explanation
$${{3{x^2} - 9x + 17} \over {{x^2} + 3x + 10}} = {{5{x^2} - 7x + 19} \over {3{x^2} + 5x + 12}}$$
$$ \Rightarrow {{3{x^2} - 9x + 17} \over {5{x^2} - 7x + 19}} = {{{x^2} + 3x + 10} \over {3{x^2} + 5x + 12}}$$
$${{ - 2{x^2} - 2x - 2} \over {5{x^2} - 7x + 19}} = {{ - 2{x^2} - 2x - 2} \over {3{x^2} + 5x + 12}}$$
Either $${x^2} + x + 1 = 0$$ or No real roots
$$ \Rightarrow 5{x^2} - 7x + 19 = 3{x^2} + 5x + 12$$
$$2{x^2} - 12x + 7 = 0$$
sum of roots = 6
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