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WAEC Further Mathematics 2017 Paper

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Question : 16

Total: 40

\alpha

\beta

are the roots of the equation

2x^{2} - 3x + 4 = 0

\;\frac{\alpha}{\beta} + \;\frac{\beta}{\alpha}

$\frac{-9}{8}$
$\frac{-7}{8}$
$\frac{7}{8}$
$\frac{9}{8}$

Solution:

Concept:

For a quadratic equation

ax^{2} + bx + c = 0

with roots

\alpha

and

\beta

, we have

\alpha + \beta = -\frac{b}{a}

and

\alpha\beta = \frac{c}{a}

.

Explanation:

First, rewrite the required expression as

\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^{2} + \beta^{2}}{\alpha\beta}

.

Use the identity

\alpha^{2} + \beta^{2} = (\alpha + \beta)^{2} - 2\alpha\beta

.

From

2x^{2} - 3x + 4 = 0

, we have

a = 2

,

b = -3

,

c = 4

.

So

\alpha + \beta = \frac{-(-3)}{2} = \frac{3}{2}

and

\alpha\beta = \frac{4}{2} = 2

.

Then

\alpha^{2} + \beta^{2} = \left(\frac{3}{2}\right)^{2} - 2(2) = \frac{9}{4} - 4 = -\frac{7}{4}

.

Therefore,

\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{-\frac{7}{4}}{2} = -\frac{7}{8}

.

Answer:

-\frac{7}{8}

(Option B).

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