Answers

2014-05-23T14:58:06+05:30
My understanding of the question is this:

If x = 2- \sqrt{3}  then find the value of x^2 + \frac{1}{x^2 } and x^2 - \frac{1}{x^2 }

Solution:
x = 2- \sqrt{3}
Squaring on both sides

x^2 = 4 + 3 - 4 \sqrt{3} = 7-4 \sqrt{3}

 \frac{1}{x^2} =  \frac{1}{7-4 \sqrt{3}}

On Rationalizing

\frac{1}{x^2} = \frac{1}{7-4 \sqrt{3}}.  \frac{7+4 \sqrt{3}}{7+4 \sqrt{3}}

\frac{1}{x^2} = \frac{7+4 \sqrt{3}}{49 - 48} = 7+4 \sqrt{3}

Now, 

x^2 + \frac{1}{x^2 } = 7-4 \sqrt{3} + 7+ 4 \sqrt{3} = 14

And,

x^2 - \frac{1}{x^2 } = 7-4 \sqrt{3} - 7- 4 \sqrt{3} = -8 \sqrt{3}
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