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________.
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________.
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When a rational equation has numerators and denominators whose pairwise differences turn out to be the same polynomial, subtract them first to find that common difference -- it usually splits the whole problem into two much simpler cases instead of forcing a cross-multiplied quartic.
Updated On: Aug 17, 2026
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Approach Solution - 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 \(D=1-4<0\)
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Approach Solution -2
Concept:
Step 1: Name the four polynomials and find their differences.
Let $f=3x^2-9x+17$, $g=x^2+3x+10$ (left side), and $p=5x^2-7x+19$, $q=3x^2+5x+12$ (right side), so the equation is $\frac{f}{g}=\frac{p}{q}$. Compute:
$p-f = (5x^2-7x+19)-(3x^2-9x+17) = 2x^2+2x+2$
$q-g = (3x^2+5x+12)-(x^2+3x+10) = 2x^2+2x+2$
Both differences are equal; call this common value $M = 2x^2+2x+2$.
Step 2: Substitute $p=f+M$ and $q=g+M$ into the equation.
$\frac{f}{g}=\frac{f+M}{g+M}$
Cross-multiplying: $f(g+M) = g(f+M)$, which expands to $fg+fM = fg+gM$, so $fM = gM$, giving $M(f-g)=0$.
Step 3: Solve each factor.
Case $M=0$: $2x^2+2x+2=0 \Rightarrow x^2+x+1=0$, with discriminant $1-4=-3 < 0$, so no real roots.
Case $f=g$: $3x^2-9x+17 = x^2+3x+10 \Rightarrow 2x^2-12x+7=0$, with discriminant $144-56=88>0$, so this gives two real roots.
Step 4: Find the sum of the real roots.
For $2x^2-12x+7=0$, sum of roots $= \frac{12}{2} = 6$.
Final Answer: The sum of all real values of $x$ is 6.
- When the numerator and denominator differences on both sides of a rational equation turn out to be the same polynomial, subtracting them first avoids cross-multiplying into a quartic.
Step 1: Name the four polynomials and find their differences.
Let $f=3x^2-9x+17$, $g=x^2+3x+10$ (left side), and $p=5x^2-7x+19$, $q=3x^2+5x+12$ (right side), so the equation is $\frac{f}{g}=\frac{p}{q}$. Compute:
$p-f = (5x^2-7x+19)-(3x^2-9x+17) = 2x^2+2x+2$
$q-g = (3x^2+5x+12)-(x^2+3x+10) = 2x^2+2x+2$
Both differences are equal; call this common value $M = 2x^2+2x+2$.
Step 2: Substitute $p=f+M$ and $q=g+M$ into the equation.
$\frac{f}{g}=\frac{f+M}{g+M}$
Cross-multiplying: $f(g+M) = g(f+M)$, which expands to $fg+fM = fg+gM$, so $fM = gM$, giving $M(f-g)=0$.
Step 3: Solve each factor.
Case $M=0$: $2x^2+2x+2=0 \Rightarrow x^2+x+1=0$, with discriminant $1-4=-3 < 0$, so no real roots.
Case $f=g$: $3x^2-9x+17 = x^2+3x+10 \Rightarrow 2x^2-12x+7=0$, with discriminant $144-56=88>0$, so this gives two real roots.
Step 4: Find the sum of the real roots.
For $2x^2-12x+7=0$, sum of roots $= \frac{12}{2} = 6$.
Final Answer: The sum of all real values of $x$ is 6.
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