Find the quadratic equation whose roots are 2/3 { and } -3/4

Correct Answer is : 12y2+y−6=012y² + y - 6 = 012y2+y−6=0

Correct Answer is : 12y2+y−6=012y² + y - 6 = 012y2+y−6=0

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WAEC Mathematics 2023 Paper

Question : 41

Total: 50

\frac{2}{3} \text{ and } \frac{-3}{4}

Solution:

Concept:

\alpha

and

\beta

are the roots, the quadratic equation is given by

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

Explanation:

Let

\alpha = \frac{2}{3}

and

\beta = \frac{-3}{4}

Then the equation is

\left(y - \frac{2}{3}\right)\left(y + \frac{3}{4}\right) = 0

Expanding:

y^2 + \frac{3}{4}y - \frac{2}{3}y - \frac{1}{2} = 0

Combine the

y

terms:

\frac{3}{4} - \frac{2}{3} = \frac{9 - 8}{12} = \frac{1}{12}

y^2 + \frac{1}{12}y - \frac{1}{2} = 0

Multiply through by the LCM of

3

and

4

, which is

12

12y^2 + y - 6 = 0

Answer:

The quadratic equation is

12y^2 + y - 6 = 0

Correct option: C.

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