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Surds and Indices Practice Test 2

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Question : 17 of 17

Marks: +1, -0

If

x = 5 + 2\sqrt{6}

, then find the value of

\frac{x-1}{\sqrt{x}}

.

$2\sqrt{3}$
$\sqrt{3}$
$2\sqrt{2}$
$\sqrt{2}$

Solution:

Concept:

Recognise

x

as a perfect square of a surd expression, then simplify the given fraction.

Explanation:

Given

x = 5 + 2\sqrt{6}

.

Write it as

x = 3 + 2 + 2\sqrt{3}\sqrt{2}

.

So,

x = (\sqrt{3} + \sqrt{2})^2

.

Since

x>0

,

\sqrt{x} = \sqrt{3} + \sqrt{2}

.

Now,

\frac{x-1}{\sqrt{x}} = \frac{(\sqrt{3}+\sqrt{2})^2 - 1}{\sqrt{3}+\sqrt{2}}

.

This becomes

(\sqrt{3}+\sqrt{2}) - \frac{1}{\sqrt{3}+\sqrt{2}}

.

Rationalise:

\frac{1}{\sqrt{3}+\sqrt{2}} = \sqrt{3} - \sqrt{2}

.

Thus,

\frac{x-1}{\sqrt{x}} = \sqrt{3}+\sqrt{2} - (\sqrt{3}-\sqrt{2})

.

Hence,

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

.

Answer:

2\sqrt{2}

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