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WAEC Mathematics 2020 Paper

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Question : 14

Total: 48

Find the quadratic equation whose roots are

\;\frac{1}{2}

\;\frac{1}{3}

3x $^2$ + x + 1 = 0
6x $^2$ + x - 1 = 0
3x $^2$ + x - 1 = 0
6x $^2$ - x - 1 = 0

Solution:

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

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