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WAEC Mathematics 2011 Paper

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Question : 23

Total: 47

If

\sqrt{72} + \sqrt{32} - 3 \sqrt{18} = x \sqrt{8}

, Find the value of x

1
$\; \frac{3}{4}$
$\; \frac{1}{2}$
$\; \frac{1}{4}$

Solution:

Concept:

Express each surd in terms of

\sqrt{2}

and then equate the two sides.

Explanation:

Given:

\sqrt{72} + \sqrt{32} - 3\sqrt{18} = x\sqrt{8}

.

Simplify each surd separately.

\sqrt{72} = \sqrt{36 \times 2} = 6\sqrt{2}

\sqrt{32} = \sqrt{16 \times 2} = 4\sqrt{2}

3\sqrt{18} = 3\sqrt{9 \times 2} = 3 \times 3\sqrt{2} = 9\sqrt{2}

Also,

\sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2}

.

So the left side becomes

6\sqrt{2} + 4\sqrt{2} - 9\sqrt{2}

.

Combine like terms:

(6 + 4 - 9)\sqrt{2} = \sqrt{2}

.

Thus,

\sqrt{2} = x \times 2\sqrt{2}

.

Divide both sides by

\sqrt{2}

:

1 = 2x

.

Therefore,

x = \frac{1}{2}

.

Answer:

x = \frac{1}{2}

, which is Option C.

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