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WAEC Mathematics 2010 Paper

Question : 31

Total: 49

If x2 + kx +

\frac{16}{9}

is a perfect square, find the value of k

Solution:

Concept:

For a quadratic expression to be a perfect square, its discriminant must be zero.

Explanation:

Compare

x^2 + kx + \frac{16}{9}

with

ax^2 + bx + c

Here,

a = 1

b = k

, and

c = \frac{16}{9}

A perfect square has discriminant

b^2 - 4ac = 0

So,

k^2 - 4(1)\left(\frac{16}{9}\right) = 0

This gives

k^2 - \frac{64}{9} = 0

Therefore,

k^2 = \frac{64}{9}

Taking the square root:

k = \pm \frac{8}{3}

Among the given options, the positive value is

\frac{8}{3}

Answer:

k = \frac{8}{3}

, i.e. Option A.

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