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WAEC 2020 Mathematics · Question 41

Find the quadratic equation whose roots are 12\frac{1}{2}  and -13\frac{1}{3}

Options

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

Answer: D. 6x2^2 - x - 1 = 0

Explanation written with AI and checked against the answer key

Analyze the question: We need a quadratic equation whose roots (solutions) are 1/2 and -1/3. Concept: If the roots are α and β, the equation is x² − (α + β)x + αβ = 0 (sum of roots and product of roots). Solution: Sum of roots: 1/2 + (−1/3) = 3/6 − 2/6 = 1/6 Product of roots: (1/2) × (−1/3) = −1/6 Substitute into the formula: x² − (1/6)x − 1/6 = 0 Multiply through by 6 to remove fractions: 6x² − x − 1 = 0 Check using factors: (2x − 1)(3x + 1) = 6x² + 2x − 3x − 1 = 6x² − x − 1 ✓ Tip: Quickly check any option by substituting a root. For x = 1/2 in option D: 6(1/4) − 1/2 − 1 = 1.5 − 0.5 − 1 = 0 ✓. Also, remember that the sign of the middle term is the opposite of the sum of the roots (when the leading coefficient is positive). Here the sum is positive (1/6), so the middle term is negative.

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