The sum of all real values of x for which (3 x²-9 x+17)/(x²+3 x+10)=(5 x²-7 x+19)/(3 x²+5 x+12) is equal to __________.

Solution

Maths sir · 5 steps

Subtract each numerator from the other and each denominator from the other: both give the same quadratic.

1. Spot the pattern. Call the fractions a/b = c/d. Then c - a = 2x² + 2x + 2 and d - b = 2x² + 2x + 2, so both differences equal 2(x² + x + 1).
2. Use the equality. Cross-multiplying gives ad = bc. Subtract ab from both sides to get a(d - b) = b(c - a), which here is 2(x² + x + 1)(a - b) = 0.
3. Cancel the common factor. x² + x + 1 has discriminant 1 - 4 < 0, so it is never zero for real x. Hence a = b.

   3x² - 9x + 17 = x² + 3x + 10 ⇒ 2x² - 12x + 7 = 0

4. Check the roots. The discriminant is 144 - 56 = 88 > 0, so both roots are real. Neither denominator can vanish, since 9 - 40 < 0 and 25 - 144 < 0.
5. So. The sum of the real roots is 12/2 = 6.

The common slip: 5. Adding all four roots of the cross-multiplied quartic gives 5, not 6. The factor x² + x + 1 contributes two complex roots with sum -1.

The idea to keep. When two equal fractions look messy, compare the numerator difference with the denominator difference. If the two differences match and never vanish, the fractions must both equal 1.

Answer from the NTA final answer key. Solution by PrepWiser.