Test Index

WAEC Mathematics 2020 Paper

© africaexams.com

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:

Concept:

If

\alpha

and

\beta

are roots, the quadratic equation is

(x - \alpha)(x - \beta) = 0

.

Explanation:

The roots are

\frac{1}{2}

and

-\frac{1}{3}

.

So,

\left(x - \frac{1}{2}\right)\left(x + \frac{1}{3}\right) = 0

.

Clear fractions by multiplying the factors:

(2x - 1)(3x + 1) = 0

.

Expand:

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

.

Simplify:

6x^2 - x - 1 = 0

.

Answer:

6x^2 - x - 1 = 0

, which is option D.

© africaexams.com

Go to Question:

Prev Question

Next Question