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WAEC 2023 Mathematics · Question 50

Find the quadratic equation whose roots are 23and−34\frac{2}{3} and \frac{- 3}{4}

Options

  1. A.12y2−y−6=012y^2 - y - 6 = 0
  2. B.12y2−y+6=012y^2 - y + 6 = 0
  3. C.12y2+y−6=012y^2 + y - 6 = 0
  4. D.y2+y−6=0y^2 + y - 6 = 0

Answer: C. 12y2+y−6=012y^2 + y - 6 = 0

Explanation written with AI and checked against the answer key

Step 1: Analyze the question We need a quadratic equation whose roots are 2/3 and -3/4. Step 2: The concept If the roots are α and β, the equation is: y² - (α + β)y + αβ = 0 Step 3: Solution Sum of roots: α + β = 2/3 + (-3/4) = 8/12 - 9/12 = -1/12 Product of roots: αβ = (2/3)(-3/4) = -6/12 = -1/2 Substitute into the formula: y² - (-1/12)y + (-1/2) = 0 y² + (1/12)y - 1/2 = 0 Multiply through by 12 to clear the fractions: 12y² + y - 6 = 0 Check using factors: (3y - 2)(4y + 3) = 12y² + 9y - 8y - 6 = 12y² + y - 6 ✓ Step 4: Answer The equation is 12y² + y - 6 = 0, which is option C. Tip: Always use "y² − (sum)y + product = 0". Watch the signs. A negative sum becomes a positive middle term. Also, you can quickly eliminate options: option D has no 12 coefficient, so it can't give fractional roots like these, and the product of roots must be negative, which rules out B.

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