WAEC - Mathematics (2019 - No. 13)

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

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\(\frac{1}{2}\) 

Comments (0)

Advertisement