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 (integer)
6
Solution
<p>$${{3{x^2} - 9x + 17} \over {{x^2} + 3x + 10}} = {{5{x^2} - 7x + 19} \over {3{x^2} + 5x + 12}}$$</p>
<p>$$ \Rightarrow {{3{x^2} - 9x + 17} \over {5{x^2} - 7x + 19}} = {{{x^2} + 3x + 10} \over {3{x^2} + 5x + 12}}$$</p>
<p>$${{ - 2{x^2} - 2x - 2} \over {5{x^2} - 7x + 19}} = {{ - 2{x^2} - 2x - 2} \over {3{x^2} + 5x + 12}}$$</p>
<p>Either ${x^2} + x + 1 = 0$ or No real roots</p>
<p>$\Rightarrow 5{x^2} - 7x + 19 = 3{x^2} + 5x + 12$</p>
<p>$2{x^2} - 12x + 7 = 0$</p>
<p>sum of roots = 6</p>
About this question
Subject: Mathematics · Chapter: Complex Numbers and Quadratic Equations · Topic: Complex Numbers and Argand Plane
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