WAEC - Mathematics (2020 - No. 14)
Find the quadratic equation whose roots are \(\frac{1}{2}\) and -\(\frac{1}{3}\)
3x\(^2\) + x + 1 = 0
6x\(^2\) + x - 1 = 0
3x\(^2\) + x - 1 = 0
6x\(^2\) - x - 1 = 0
Giải thích
x = \(\frac{1}{2}\) and x = \(\frac{-1}{3}\)
(2x - 1) = 0 and (3x + 1) = 0
(2x - 1) (3x + 1) = 0
6x\(^2\) - x - 1 = 0