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________.

Hint:

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.

Answer 1

Given :

\(\begin{array}{l} \frac{3x^2-9x+17}{x^2+3x+10}=\frac{5x^2-7x+19}{3x^2+5x+12} \end{array}\)

Cross multiply

\(\begin{array}{l} (3x^2-9x+17)(3x^2+5x+12)=(5x^2-7x+19)(x^2+3x+10) \end{array}\)

Expand LHS

\(\begin{array}{l} = 9x^4+15x^3+36x^2 -27x^3-45x^2-108x +51x^2+85x+204 \end{array}\)

\(\begin{array}{l} = 9x^4 -12x^3 +42x^2 -23x +204 \end{array}\)

Expand RHS

\(\begin{array}{l} = 5x^4+15x^3+50x^2 -7x^3-21x^2-70x +19x^2+57x+190 \end{array}\)

\(\begin{array}{l} = 5x^4 +8x^3 +48x^2 -13x +190 \end{array}\)

Equate LHS = RHS

\(\begin{array}{l} 9x^4 -12x^3 +42x^2 -23x +204 = 5x^4 +8x^3 +48x^2 -13x +190 \end{array}\)

\(\begin{array}{l} 4x^4 -20x^3 -6x^2 -10x +14 = 0 \end{array}\)

Divide by 2

\(\begin{array}{l} 2x^4 -10x^3 -3x^2 -5x +7 = 0 \end{array}\)

Factorization

\(\begin{array}{l} ( x-1 )( 2x^3 -8x^2 -11x -7 ) = 0 \end{array}\)

From this, real root is \(x=1\)

The cubic gives one real root \(x=7\)

Sum of real roots

\(\begin{array}{l} = 1 + 7 = 8 \end{array}\)

Hence, answer = 8