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WAEC 2017 Mathematics · Question 40

The roots of a quadratic equation are −12\frac{-1}{2} and 23\frac{2}{3}. Find the equation.

Options

  1. A.6x2−x+2=06x^2 - x + 2 = 0
  2. B.6x2−x−2=06x^2 - x - 2 = 0
  3. C.6x2+x−2=06x^2 + x - 2 = 0
  4. D.6x2+x+2=06x^2 + x + 2 = 0

Answer: B. 6x2−x−2=06x^2 - x - 2 = 0

Explanation written with AI and checked against the answer key

Analyze the question: We are given the two roots of a quadratic equation and must find the equation. Concept: If the roots are α and β, the equation is x² − (α + β)x + αβ = 0 Solution: Sum of roots: α + β = −1/2 + 2/3 = (−3 + 4)/6 = 1/6 Product of roots: αβ = (−1/2)(2/3) = −2/6 = −1/3 Substitute into the formula: x² − (1/6)x − 1/3 = 0 Multiply through by 6 to clear the fractions: 6x² − x − 2 = 0 Check using factors: x = −1/2 gives (2x + 1) = 0 x = 2/3 gives (3x − 2) = 0 (2x + 1)(3x − 2) = 6x² − 4x + 3x − 2 = 6x² − x − 2 ✓ Answer: Option B, 6x² − x − 2 = 0. Tip: Be careful with signs. The middle term is −(sum of roots), and the constant term is the product of roots. A quick check: the product is negative (one root is positive, one negative), so the constant must be negative. That rules out A and D immediately. Then the sum is positive (1/6), so the x-term must be negative, which leaves B.

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