1.

If x 9-4root 5 find value of x2-1/x2

Answer» HEY mate!

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GIVEN :

x = 9 - 4 \sqrt{5}

To FIND :

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

Solution :

x = 9 - 4 \sqrt{5} \\ \\ \frac{1}{x} = \frac{1}{9 - 4 \sqrt{5} } \times \frac{9 + 4 \sqrt{5} }{9 + 4 \sqrt{5} } \\ \\ \frac{1}{x} = \frac{9 + 4 \sqrt{5} }{(9) {}^{2} - (4 \sqrt{5}) {}^{2} } \\ \\ \frac{1}{x} = \frac{9 + 4 \sqrt{5} }{81 - 80} \\ \\ \frac{1}{x} = 9 + 4 \sqrt{5}

Now,

x {}^{2} = (9 - 4 \sqrt{5} ) {}^{2} \\ \\ x {}^{2} = (9) {}^{2} + (4 \sqrt{5}) {}^{2} - 2 \times 9 \times 4 \sqrt{5} \\ \\ x {}^{2} = 81 + 80 - 72 \sqrt{5} \\ \\ x {}^{2} =16 1 - 72 \sqrt{5}

And,

( \frac{1}{x} ) {}^{2} = (9 + 4 \sqrt{5} ) {}^{2} \\ \\ \frac{1}{x {}^{2} } =( 9 ){}^{2} + (4 \sqrt{5} ) {}^{2} + 2 \times 2 \times 4 \sqrt{5} \\ \\ \frac{1}{x {}^{2} } = 81 + 80 + 72 \sqrt{5} \\ \\ \frac{1}{x {}^{2} } = 161 + 72 \sqrt{5}

FINALLY,

x {}^{2} - \frac{1}{x {}^{2} } \\ \\ \implies 161 - 72 \sqrt{5} - (161 + 72 \sqrt{5} ) \\ \\ \implies \cancel{161} - 72 \sqrt{5} - \cancel{161} \: - 72 \sqrt{5} \\ \\ \implies - 144 \sqrt{5}

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Thanks for the question!

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