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The sum of real roots of $\dfrac{{3{x^2} - 9x + 17}}{{{x^2} + 3x + 10}} = \dfrac{{5{x^2} - 7x + 19}}{{3{x^2} + 5x + 12}}$ is____

How do I proceed with this type of problem.

Cross multiplying and segregating the coefficient is a cumbersome process

My putting in desmos.com my answer is $«6»$.

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    $\begingroup$ Did you mean "real roots"? $\endgroup$
    – User
    Aug 5 at 16:33
  • $\begingroup$ In desmos I could figure out only two real roots therefore two roots are imaginary $\endgroup$
    – Samar Imam Zaidi
    Aug 5 at 16:34
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    $\begingroup$ Cross-multiplying leads to a 4th degree equation that can be factored as $2(x^2 + x + 1)(2x^2 - 12x + 7).$ FYI, I don't follow @User's suggestion, since I don't see how $ad = bc$ is equivalent to $bc - bd = ac - bc.$ (moments later) OK, the comment was omitted and replaced with an answer, and the answer gives a different fraction equation than the original comment. $\endgroup$
    – Dave L. Renfro
    Aug 5 at 16:49

2 Answers 2

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Suggestion (Hint):

Use the well-known rule:

$$\frac ab=\frac cd \iff \frac {a-b}{b}=\frac {c-d}{d}$$

Then observe that: $$a-b=c-d$$

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    $\begingroup$ Very nice! Once you see it, it's obvious. But how did you see it? $\endgroup$
    – TonyK
    Aug 5 at 16:55
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    $\begingroup$ (+1) OK, this works. Neat trick, although it seems like it is one of those "only works on very select types of math competition problems". Incidentally, this only gives two solutions (even over the complex numbers), but in thinking about it, cross-multiplying as I did in a comment could introduce one or more extraneous solutions by multiplying both sides by variable expressions that could equal zero. $\endgroup$
    – Dave L. Renfro
    Aug 5 at 16:57
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    $\begingroup$ @DaveL.Renfro Following User's approach you should still get four roots in $\mathbb C$, two of which are real. Two real roots from $a-b=c-d=0$, and two roots from setting $b=d$ which happen to be non-real cube roots of unity... $\endgroup$
    – Macavity
    Aug 5 at 17:12
  • $\begingroup$ @TonyK But, $a-b=0$ (or $c-d=0$) gives us a quadratic equation. Did I get your point wrong? $\endgroup$
    – User
    Aug 9 at 11:50
  • $\begingroup$ @User: You are right. I misread Macavity's comment, so I am deleting mine. $\endgroup$
    – TonyK
    Aug 9 at 16:37
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Cross-multiplying gives:

$$(3x^2 - 9x + 17)(3x^2 + 5x + 12) = (x^2 + 3x + 10)(5x^2 - 7x + 19)$$ $$9x^4 + 15x^3 + 36x^2 - 27x^3 - 45x^2 - 108x + 51x^2 + 85x + 204 = 5x^4 - 7x^3 + 19x^2 + 15x^3 - 21x^2 + 57x + 50x^2 - 70x + 190$$ $$9x^4 - 12x^3 + 42x^2 - 23x + 204 = 5x^4 + 8x^3 + 48x^2 - 13x + 190$$ $$4x^4 - 20x^3 - 6x^2 - 10x + 14 = 0$$ $$x^4 - 5x^3 - \frac{3}{2}x^2 - \frac{5}{2}x + \frac{7}{2} = 0$$

There, that wasn't too bad. Now the hard part: finding the roots.

The Rational Root Theorem tells us that all rational roots are in $\pm\lbrace 1, 7, \frac{1}{2}, \frac{7}{2} \rbrace$. Unfortunately, it turns out that none of those actually are roots, so let's take another path and try a quadratic factorization:

$$f(x) = (x^2 + Ax + B)\left(x^2 - (A + 5)x + \frac{7}{2B}\right)$$

where the coefficients in the second factor are set to make the cubic ($-5$) and constant ($\frac{7}{2}$) coefficients correct. Expanding the product:

$$f(x) = x^4 - (A + 5)x^3 + \frac{7}{2B}x^2 + Ax^3 - A(A + 5)x^2 + \frac{7A}{2B}x + Bx^2 - (A + 5)Bx + \frac{7}{2}$$ $$f(x) = x^4 - 5x^3 + \left(\frac{7}{2B} - A^2 - 5A + B \right)x^2 + \left(\frac{7A}{2B} - AB - 5B\right)x + \frac{7}{2}$$

Solving for the $x^2$ and $x$ coefficients is a bit difficult, but it turns out that $A = B = 1$ works, so:

$$f(x) = (x^2 + x + 1)\left(x^2 - 6x + \frac{7}{2}\right) = 0$$

The first factor has a negative discriminant (-3) and thus no real solutions, but the second one has a positive discriminant (22).

We don't actually have to find the roots, just use the fact that the two roots of a quadratic $ax^2 + bx + c = 0$ add up to $\frac{-b}{a}$.

So the answer to your question is indeed 6.

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    $\begingroup$ @Macavity: Good point. I'll have to edit my answer. $\endgroup$
    – Dan
    Aug 5 at 18:22

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