Question

If x =

$(3 \: + \: \sqrt{8} )$

Show that

$( \: {x}^{2} + \: \frac{1}{ {x}^{2} } \: ) \: \: = 34$

Answer 1

★Ello ★

here is your answer !!


x = 3 + ✓8


there fore ,

1 /x = 1 / 3 + √ 8 × 3 -✓8 / 3 - ✓ 8

{ rationalize. }


1 / x = 3 -✓8 / 9 -8

=> 1 / x = 3 -✓ 8 / 1


now ,.



x + 1 / x => 3 + √ 8 + 3 - ✓ 8

=> x + 1 / x = 6



=> (x + 1 / x )² = x² + 1 / x² + 2 × x × 1 / x


=> ( 6 )² = x² + 1 / x² + 2


x² + 1 / x² = 36 -2


=> x² + 1/ x² = 34 .



hence proved !!!



hope it helps you dear !!

thanks !!

$thanks$

Answer 2

Answer:

given,

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

now,

$x + \frac{1}{x} \\ = (3 + \sqrt{8} ) + (3 - \sqrt{8} ) \\ = 3 + \sqrt{8} + 3 - \sqrt{8} \\ = 6$

finally,

${x}^{2} + \frac{1}{ {x}^{2} } \\ = (x + \frac{1}{x} ) {}^{2} - 2 \times x \times \frac{1}{x} \\ = (6) {}^{2} - 2 \\ = 36 - 2 \\ = 34$

hence proved