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}−1/\x^{2}?

4√3
6√3
4
8√3

Solution:

Given:

\x=2+\√{3}

Concept Used:
If

\x=\a+\√{\b}

, then

1 / \x=\a-\√{\b}

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

Calculation:

\x=2+\√{3}

{\x}={1}/{(2+\√{3})}

⇒{1}/{\x}={2-\√{3}}/{(2-\√{3}) \×(2+\√{3})}

\⇒{1}/{\x}={2-\√{3}}/{4-3}=2-\√{3}

(\x+1 / \x)=(2+\√{3})+(2-\√{3}) =4

And,

(\x-1 / \x)=(2+\√{3})-(2-\√{3})=2 \√{3}

\x^{2}-{1}/{\x^{2}}=(\x+1 / \x)(\x-1 / \x)

\⇒\x ^{2}-{1}/{ \x ^{2}}=4 \×2√{3}=8 \√{3}

\∴

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

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