Question

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

Answer 1

ʜᴇʟʟᴏ ᴍᴀᴛᴇ!

x = 9 - 4√5

1/x = 1/( 9 - 4√5 )

Rationalize

1/( 9 - 4√5 ) × ( 9 + 4√5 )/( 9 + 4√5 )

( 9 + 4√5 )/[ 9² - (4√5)² ] = ( 9 + 4√5 )/( 81 - 80 )

= 9 + 4√5

x² = ( 9 - 4√5 )² = 81 + 80 - 72√5

(1/x)² = ( 9 + 4√5 )² = 81 + 80 + 72√5

161 - 72√5 - 161 - 72√5 = - 144√5

Or

x² - (1/x)² = ( x + 1/x )( x - 1/x )

( 9 - 4√5 + 9 + 4√5 )[ 9 - 4√5 - ( 9 + 4√5 ) ]

( 18 ) ( - 8√5 ) = - 144√5

Hope it helps

Answer 2

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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