Question

If x=3+sqrt of 8 , find x^4+1/x^4

Answer 1

x = 3 +√8, 1/x = 3- √8


x^4 + 1/x^4
= (x² + 1/x²)²- 2
= ( (x+1/x)² - 2 )² - 2
= (6² - 2)² -2
= 34² - 2
= 1156 -2
= 1154

Answer 2

Hey friend,
Here is the answer you were looking for:
$x = 3 + \sqrt{8} \\ \\ x = 3 + \sqrt{2 \times 2 \times 2} \\ \\ x = 3 + 2 \sqrt{2} \\ \\ \frac{1}{x} = \frac{1}{3 + 2 \sqrt{2} } \\ \\ \frac{1}{x} = \frac{1}{3 + 2 \sqrt{2} } \times \frac{3 - 2 \sqrt{2} }{3 - 2 \sqrt{2} } \\ \\ using \: the \: identity \\ (a + b)(a - b) = {a}^{2} - {b}^{2} \\ \\ = \frac{3 - 2 \sqrt{2} }{ {(3)}^{2} - {(2 \sqrt{2} )}^{2} } \\ \\ \frac{1}{x} = \frac{3 - 2 \sqrt{2} }{9 - 8} \\ \\ \frac{1}{x} = 3 - 2 \sqrt{2} \\ \\ x + \frac{1}{x} = (3 + 2 \sqrt{2} ) + (3 - 2 \sqrt{2} ) \\ \\ x + \frac{1}{x} = 3 + 2 \sqrt{2} + 3 - 2 \sqrt{2} \\ \\ x + \frac{1}{x} = 3 + 3 \\ \\ x + \frac{1}{x} = 6 \\ \\ {(x + \frac{1}{x}) }^{2} = {(6)}^{2} \\ \\ {(x)}^{2} + {( \frac{1}{x} )}^{2} + 2 \times x \times \frac{1}{ x } = 36 \\ \\ {x}^{2} + \frac{1}{ {x}^{2} } + 2 = 36 \\ \\ {x}^{2} + \frac{1}{ {x}^{2} } = 36 - 2 \\ \\ {x}^{2} + \frac{1}{ {x}^{2} } = 34 \\ \\ {( {x}^{2} + \frac{1}{ {x}^{2} } )}^{2} = {(34)}^{2} \\ \\ {( {x}^{2} )}^{2} + {( \frac{1}{ {x}^{2} } )}^{2} + 2 \times {x}^{2} \times \frac{1}{ {x}^{2} } = 1156 \\ \\ {x}^{4} + \frac{1}{ {x}^{4} } + 2 = 1156 \\ \\ {x}^{4} + \frac{1}{ {x}^{4} } = 1156 - 2 \\ \\ {x}^{4} + \frac{1}{ {x}^{4} } = 1154$


Hope this helps!!!!

@Mahak24

Thanks...
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