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Haryana Police Constable Model Paper 1

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Question : 49 of 100

Marks: +1, -0

If x = 2 + √3, then what is the value of

x^{2}-\frac{1}{x^{2}}?

4√3
6√3
4
8√3

Solution:

Given:

x=2+\sqrt{3}

Concept Used:

If

x=a+\sqrt{b}

, then

\frac{1}{x}=a-\sqrt{b}

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

Calculation:

x=2+\sqrt{3}

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

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

\Rightarrow\frac{1}{x}=\frac{2-\sqrt{3}}{4-3}=2-\sqrt{3}

\left(x+\frac{1}{x}\right)=(2+\sqrt{3})+(2-\sqrt{3})=4

And,

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

x^{2}-\frac{1}{x^{2}}=\left(x+\frac{1}{x}\right)\left(x-\frac{1}{x}\right)

\Rightarrow x^{2}-\frac{1}{x^{2}}=4\times 2\sqrt{3}=8\sqrt{3}

\therefore

The required value of $x^{2}-\frac{1}{x^{2}}$ is $8\sqrt{3}$ .

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