Question

If x=Root of 3-2root2 find x+1/x and x square +1/x square

Answer 1

Solution:

Given that ;

x = √3 - 2√2

Find the value of 1/x.

$\implies \sf \frac{1}{x} = \frac{1}{ \sqrt{3} - 2 \sqrt{2} } \\ \\ \implies \sf \frac{1}{x} = \frac{1}{3 - 2 \sqrt{2} } \times \frac{ \sqrt{3} + 2 \sqrt{2} }{ \sqrt{3} + 2 \sqrt{2} } \\ \\ \implies \sf \frac{1}{x} = \frac{ \sqrt{3} + 2 \sqrt{2} }{( \sqrt{3}) {}^{2} - (2 \sqrt{2}) {}^{2} } \\ \\ \implies \sf \frac{1}{x} = \frac{ \sqrt{3} + 2 \sqrt{2} }{3 - 8} \\ \\ \implies \sf \frac{1}{x} = \frac{ \sqrt{3} + 2 \sqrt{2} }{ - 5} \\ \\ \implies \sf \frac{1}{x} = \frac{2 \sqrt{2} - \sqrt{3} }{5}$

Now, find x + (1/x),

$\implies \sf x + \frac{1}{x} = \sqrt{3} - 2 \sqrt{2} + \frac{2 \sqrt{2} - \sqrt{3} }{5} \\ \\ \implies \sf x + \frac{1}{x} = \frac{5 \sqrt{3} - 10 \sqrt{2} + 2 \sqrt{2} - \sqrt{3} }{5} \\ \\ \implies \sf x + \frac{1}{x} = \frac{4 \sqrt{3} - 8 \sqrt{2} }{5}$

Squaring both sides ;

$\implies \sf (x + \frac{1}{x}) {}^{2} = \frac{(4 \sqrt{3} - 8 \sqrt{2}) {}^{2} }{(5)^{2} } \\ \\ \implies \sf {x}^{2} + \frac{1}{ {x}^{2} } + 2 = \frac{48 + 128 - 64 \sqrt{6} }{25} \\ \\ \implies \sf {x}^{2} + \frac{1}{ {x}^{2} } = \frac{176 - 64 \sqrt{6} }{25} - 2 \\ \\ \implies \sf {x}^{2} + \frac{1}{ {x}^{2} } = \frac{176 - 64 \sqrt{6} - 50 } {25} \\ \\ \implies \sf {x}^{2} + \frac{1}{ {x}^{2} } = \frac{126 - 64 \sqrt{6} }{25}$