Test Index

WAEC Further Mathematics 2013 Paper

© africaexams.com

Question : 8

Total: 40

If

\alpha

eta

are the roots of the equation

2x^{2} - 6x + 5 = 0

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

.

$\; \frac{24}{5}$
$\; \frac{8}{5}$
$\; \frac{5}{8}$
$\; \frac{5}{24}$

Solution:

Concept:

Use the sum and product of roots of a quadratic equation to simplify the given expression.

Explanation:

For

2x^{2} - 6x + 5 = 0

, compare with

ax^{2} + bx + c = 0

.

So

a = 2

,

b = -6

,

c = 5

.

Sum of roots:

\alpha + \beta = \frac{-b}{a} = \frac{-(-6)}{2} = 3

.

Product of roots:

\alpha\beta = \frac{c}{a} = \frac{5}{2}

.

Now

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

.

Using

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

:

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

.

Substitute the values:

\frac{3^{2} - 2\left(\frac{5}{2}\right)}{\frac{5}{2}} = \frac{9 - 5}{\frac{5}{2}} = \frac{4}{\frac{5}{2}} = \frac{8}{5}

.

Answer:

\frac{8}{5}

, which is option B.

© africaexams.com

Go to Question:

Prev Question

Next Question