Question

If x= √3+√2/√3-√2 find (1) x^2 +1/x^2
(2) x^4+1/x^4

Answer 1

$\sf{x = \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } }$

rationalise the denominator

$\sf{x = \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \times \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} + \sqrt{2} } }$

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

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

$\sf{x = \frac{ 5+ 2 \sqrt{6} }{ 1 }}$

now

$\sf{ \frac{1}{x} = \frac{1}{5 + 2 \sqrt{6} } }$

rationalise the denominator

$\sf{ \frac{1}{x} = \frac{1}{5 + 2 \sqrt{6} } \times \frac{5 - 2 \sqrt{6} }{5 - 2 \sqrt{6} } }$

$\sf{ \frac{1}{x} = \frac{5 - 2 \sqrt{6} }{ {5}^{2} - {( 2 \sqrt{6} ) }^{2} } }$

$\sf{ \frac{1}{x} = \frac{5 - 2 \sqrt{6} }{ 25 - 24} }$

$\sf{ \frac{1}{x} = {5 - 2 \sqrt{6} }}$

$\sf{x + \frac{1}{x} = 5 + 2 \sqrt{6} + 5 - 2 \sqrt{6} }$

$\sf{x + \frac{1}{x} = 10}$

squaring on both sides

$1) \: \sf{{(x + \frac{1}{x})}^{2} = {10}^{2} }$

$\sf{{{x}^{2} + \frac{1}{{x}^{2} }} + 2= {100 }}$

$\fbox{\sf{{{x}^{2} + \frac{1}{{x}^{2} }} = {98 }}}$

now again squaring on both sides

$2) \: {\sf({{{{x}^{2} + \frac{1}{{x}^{2} }} )}^{2} = { {98}^{2}}}}$

${\sf{{{x}^{4} + \frac{1}{{x}^{4} }} + 2 = { 9604}}}$

$\fbox{\sf{{{x}^{4} + \frac{1}{{x}^{4} }} = { 9602}}}$

Answer 2

Answer:

Solve 1: x² + 1/x² = 98

Solve 2: (x² + 1/x²)² = 9602

Step-by-step explanation:

Refer to the attachment^^