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Q.

If $x = 3 + 2 \sqrt{2}$ , then find the value of $\sqrt{x} - \frac{1}{\sqrt{x}}$ is:

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a

\pm 2 \sqrt{2}

b

\pm 2

c

\pm \sqrt{2}

d

\pm \frac{1}{2}

answer is B.

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Detailed Solution

Concept: We will assume

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

as , i.e.

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

, then we will square on both sides of this expression, and solve it further using identity.
Let

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

be t, i.e.

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

Now, on squaring on both the sides, we get

\Rightarrow (\sqrt{x} - \frac{1}{\sqrt{x}})^{2} = t^{2}

\Rightarrow x + \frac{1}{x} - 2 = t^{2}

From here, we will substitute the value of

x

to

3 + 2 \sqrt{2}

, we get

\Rightarrow 3 + 2 \sqrt{2} + \frac{1}{3 + 2 \sqrt{2}} - 2 = t^{2}

We will now multiply and divide the given rational number by the conjugate of

3 + 2 \sqrt{2}

which is 3-2

\sqrt{2}

to rationalize the

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

\Rightarrow 3 + 2 \sqrt{2} + \frac{3 - 2 \sqrt{2}}{(3 + 2 \sqrt{2}) (3 - 2 \sqrt{2})} - 2 = t^{2}

Using the identity

(a + b) (a - b) = a^{2} - b^{2},

\Rightarrow 3 + 2 \sqrt{2} + \frac{3 - 2 \sqrt{2}}{(3)^{2} - (2 \sqrt{2})^{2}} - 2 = t^{2}

\Rightarrow 3 + 2 \sqrt{2} + \frac{3 - 2 \sqrt{2}}{9 - 8} - 2 = t^{2}

\Rightarrow 3 + 2 \sqrt{2} + 3 - 2 \sqrt{2} - 2 = t^{2}

\Rightarrow 6 - 2 = t^{2}

\Rightarrow 4 = t^{2}

\Rightarrow \pm 2 = t

The value of

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

.
Hence, the correct answer is option 2.

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