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WAEC Further Mathematics 2019 Paper

Question : 4

Total: 39

Rationalize;

\;\frac{1}{\sqrt{2\;+\;1}}

Solution:

Concept:

Rationalisation removes the radical from the denominator by multiplying with its conjugate.

Explanation:

Given expression:

\frac{1}{\sqrt{2} + 1}

Multiply numerator and denominator by the conjugate

\sqrt{2} - 1

\frac{1}{\sqrt{2} + 1} \times \frac{\sqrt{2} - 1}{\sqrt{2} - 1}

This is valid because the multiplier equals

1

In the denominator, use the identity

(a+b)(a-b)=a^2 - b^2

(\sqrt{2} + 1)(\sqrt{2} - 1) = (\sqrt{2})^2 - (1)^2 = 2 - 1 = 1

So the expression becomes

\frac{\sqrt{2} - 1}{1}

Therefore,

\frac{1}{\sqrt{2} + 1} = \sqrt{2} - 1

Answer:

\sqrt{2} - 1

, which is Option A.

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