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Q.

If $x = 3 - 2 \sqrt{2} then find x^{4} - \frac{1}{x^{4}}$

Subtopics: The Fundamental Theorem of Arithmetic

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a

- 816 \sqrt{2}

b

- 2 \sqrt{2}

c

30 \sqrt{2}

d

756 \sqrt{2}

answer is A.

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Detailed Solution

x = 3 - 2 \sqrt{2}

x^{2} = {(3 - 2 \sqrt{2})}^{2}

x^{2} = {3^{2} - 2 (3) (2 \sqrt{2}) + (2 \sqrt{2})}^{2}

x^{2} = 9 - 12 \sqrt{2} + 8

x^{2} = 17 - 12 \sqrt{2}

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

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

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

{(\frac{1}{x})}^{2} = {(3 + 2 \sqrt{2})}^{2}

{(\frac{1}{x})}^{2} = {3^{2} + 2 (3) (2 \sqrt{2}) + (2 \sqrt{2})}^{2}

{(\frac{1}{x})}^{2} = 9 + 12 \sqrt{2} + 8

{(\frac{1}{x})}^{2} = 17 + 12 \sqrt{2}

\frac{1}{x^{2}} = 17 + 12 \sqrt{2}

x^{4} - \frac{1}{x^{4}} = {(x^{2})}^{2} - \frac{1}{{(x^{2})}^{2}}

x^{4} - \frac{1}{x^{4}} = (x^{2} + \frac{1}{x^{2}}) (x^{2} - \frac{1}{x^{2}})

x^{4} - \frac{1}{x^{4}} = (17 - 12 \sqrt{2} + 17 + 12 \sqrt{2}) (17 - 12 \sqrt{2} - (17 + 12 \sqrt{2}))

x^{4} - \frac{1}{x^{4}} = (34) (17 - 12 \sqrt{2} - 17 - 12 \sqrt{2})

x^{4} - \frac{1}{x^{4}} = (34) (- 24 \sqrt{2})

x^{4} - \frac{1}{x^{4}} = - 816 \sqrt{2}

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