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\(\frac{1}{2}\)
waec 2019mathematics

The correct answer is B. x = 1\(\frac{1}{2}\) or x = 2\(\frac{1}{2}\)

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

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

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

Previous question Next question