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Solve \(4x^{2}\) – 16x + 15 = 0.

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Mathematics WAEC 2019

Solve \(4x^{2}\) – 16x + 15 = 0.

  • x = 1\(\frac{1}{2}\) or x = -2\(\frac{1}{2}\)
  • x = 1\(\frac{1}{2}\) or x = 2\(\frac{1}{2}\) checkmark
  • x = 1\(\frac{1}{2}\) or x = -1\(\frac{1}{2}\)
  • x = -1\(\frac{1}{2}\) or x -2\(\frac{1}{2}\)

The correct answer is: 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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