Join WhatsApp Icon JEE WhatsApp Group
Question 81

The sum of all real values 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


Correct Answer: 6

We need to solve $$\frac{3x^2 - 9x + 17}{x^2 + 3x + 10} = \frac{5x^2 - 7x + 19}{3x^2 + 5x + 12}$$.

First, cross-multiplying yields $$(3x^2 - 9x + 17)(3x^2 + 5x + 12) = (5x^2 - 7x + 19)(x^2 + 3x + 10)$$.

Next, we look for a common structure by introducing $$A = x^2 + 3x + 10$$ and $$B = 3x^2 + 5x + 12$$.

Computing the first numerator minus the first denominator, we find $$(3x^2 - 9x + 17) - (x^2 + 3x + 10) = 2x^2 - 12x + 7$$.

Similarly, computing the second numerator minus the second denominator gives $$(5x^2 - 7x + 19) - (3x^2 + 5x + 12) = 2x^2 - 12x + 7$$.

Since both differences equal the same expression, we let $$k = 2x^2 - 12x + 7$$.

Therefore the first numerator can be written as $$A + k$$ and the second numerator as $$B + k$$.

Substituting these expressions into the equation yields $$\frac{A + k}{A} = \frac{B + k}{B}$$.

From this we obtain $$1 + \frac{k}{A} = 1 + \frac{k}{B}$$, which simplifies to $$\frac{k}{A} = \frac{k}{B}$$ and hence $$k\left(\frac{1}{A} - \frac{1}{B}\right) = 0$$.

This gives us two possibilities: either $$k = 0$$ or $$A = B$$.

If $$k = 0$$ then $$2x^2 - 12x + 7 = 0$$. By the quadratic formula we have $$x = \frac{12 \pm \sqrt{144 - 56}}{4} = \frac{12 \pm \sqrt{88}}{4}$$, and both roots are real. By Vieta's formulas, the sum of the roots is $$\frac{12}{2} = 6$$. We must also verify these roots do not make the denominators zero: since $$A = x^2 + 3x + 10$$ has discriminant $$9 - 40 = -31 < 0$$, it is always positive, and similarly $$B = 3x^2 + 5x + 12$$ has discriminant $$25 - 144 = -119 < 0$$ and is always positive, so neither denominator vanishes. Hence both roots are valid.

On the other hand, if $$A = B$$ then $$x^2 + 3x + 10 = 3x^2 + 5x + 12$$, which simplifies to $$2x^2 + 2x + 2 = 0$$ or $$x^2 + x + 1 = 0$$. Since the discriminant $$1 - 4 = -3 < 0$$, there are no real roots in this case.

Therefore, the only real solutions come from $$k = 0$$, and their sum is $$6$$.

The answer is 6.

Get AI Help

Create a FREE account and get:

  • Free JEE Mains Previous Papers PDF
  • Take JEE Mains paper tests

50,000+ JEE Students Trusted Our Score Calculator

Predict your JEE Main percentile, rank & performance in seconds

Ask AI

Ask our AI anything

AI can make mistakes. Please verify important information.