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Solve \(4x^{2}\) - 16x + 15 = 0.

Mathematics
WAEC 2019

Solve \(4x^{2}\) - 16x + 15 = 0.

  • A. x = 1\(\frac{1}{2}\) or x = -2\(\frac{1}{2}\)
  • B. x = 1\(\frac{1}{2}\) or x = 2\(\frac{1}{2}\)
  • C. x = 1\(\frac{1}{2}\) or x = -1\(\frac{1}{2}\)
  • D. x = -1\(\frac{1}{2}\) or x -2\(\frac{1}{2}\)
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Correct Answer: Option B
Explanation

4x\(^2\) - 16x + 15 = 0

\(4x^2\) - 6x - 10x + 15 = 0

2x(2x - 3) - 5(2x - 3) = 0

(2x - 3)(2x - 5) = 0

x = 1\(\frac{1}{2}\) or x = 2\(\frac{1}{2}\) 


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