1.

If x=3-2√2, Find the value of x^2+1/x^2

Answer»

Answer:

GIVEN :

x = 3 - 2 \SQRT{2}

To find :

x {}^{2} + \frac{1}{x {}^{2} }

Solution :

x = 3 - 2 \sqrt{2} \\ \\ \frac{1}{x} = \frac{1}{3 - 2 \sqrt{2}} \times \frac{3 + 2 \sqrt{2}}{3 + 2 \sqrt{2}} \\ \\ \frac{1}{x} = \frac{3 + 2\sqrt{2} }{(3) {}^{2} - (2 \sqrt{2} ) {}^{2} } \\ \\ \frac{1}{x} = \frac{3 + 2 \sqrt{2} }{9 - 8} \\ \\ \frac{1}{x} = 3 + 2 \sqrt{2}

Now,

x + \frac{1}{x} = 3 - \cancel{2 \sqrt{2}} + 3 + \cancel{2 \sqrt{2}} \\ \\ x + \frac{1}{x} = 3 + 3 \\ \\ x + \frac{1}{x} = 6

Again,

\bold{on \: squaring \: both \: sides.} \\ \\ (x + \frac{1}{x}) {}^{2} = (6) {}^{2} \\ \\ x {}^{2} + \frac{1}{x {}^{2} } + 2 \times \cancel x \times \frac{1}{ \cancel x} = 36 \\ \\ x {}^{2} + \frac{1}{x {}^{2} }+ 2 = 36 \\ \\ x {}^{2} + \frac{1}{x {}^{2} } = 36 - 2 \\ \\ \THEREFORE \boxed{\bold{x {}^{2} + \frac{1}{x {}^{2} } = 34}}

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