The sum of all real values of x for which fra | vedclass

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(B) Given equation: $\frac{3 x^{2}-9 x+17}{x^{2}+3 x+10}=\frac{5 x^{2}-7 x+19}{3 x^{2}+5 x+12}$Rewrite the numerators:$\frac{(x^{2}+3 x+10) + (2 x^{2}

Vedclass · Accepted Answer

$6$

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(B) Given equation: $\frac{3 x^{2}-9 x+17}{x^{2}+3 x+10}=\frac{5 x^{2}-7 x+19}{3 x^{2}+5 x+12}$Rewrite the numerators:$\frac{(x^{2}+3 x+10) + (2 x^{2}-12 x+7)}{x^{2}+3 x+10} = \frac{(3 x^{2}+5 x+12) + (2 x^{2}-12 x+7)}{3 x^{2}+5 x+12}$$1 + \frac{2 x^{2}-12 x+7}{x^{2}+3 x+10} = 1 + \frac{2 x^{2}-12 x+7}{3 x^{2}+5 x+12}$$(2 x^{2}-12 x+7) \left( \frac{1}{x^{2}+3 x+10} - \frac{1}{3 x^{2}+5 x+12} \right) = 0$Case $1$: $2 x^{2}-12 x+7 = 0$. The sum of roots is $-\frac{b}{a} = -(\frac{-12}{2}) = 6$. The discriminant $D = (-12)^{2} - 4(2)(7) = 144 - 56 = 88 > 0$,so both roots are real.Case $2$: $x^{2}+3 x+10 = 3 x^{2}+5 x+12 \implies 2 x^{2}+2 x+2 = 0 \implies x^{2}+x+1 = 0$. The discriminant $D = 1^{2} - 4(1)(1) = -3 Thus,the sum of all real values of $x$ is $6$.

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.

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