Question

if x=3+2root2, then find the value of x^4+1/x^4

Answer 1

first you. have to find the value of x^4
that is (3+2√2)²*(3+2√2)²
by finding that you should put that value in equation
hence you will get your answer!!

Answer 2

Hey dear,
Here is the answer you were looking for:
$x = 3 + 2 \sqrt{2} \\ \\ \frac{1}{x} = \frac{1}{3 + 2 \sqrt{2} }$

On rationalizing the denominator we get,

$\frac{1}{x} = \frac{1}{3 + 2 \sqrt{2} } \times \frac{3 - 2 \sqrt{2} }{3 - 2 \sqrt{2} }$

Using the identity:

$(x + y)(x - y) = {x}^{2} - {y}^{2}$

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

Putting the values :

$= (3 + 2 \sqrt{2} ) + (3 - 2 \sqrt{2} ) \\ \\ = 3 + 2 \sqrt{2} + 3 - 2 \sqrt{2} \\ \\ = 6 \\ \\ {x}^{2} + \frac{1}{ {x}^{2} }$

Squaring (x) + (1/x) = 6 we get,

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

Again squaring (x^2) + (1/x^2) = 34 we get,

${( {x}^{2} + \frac{1}{ {x}^{2} } )}^{2} = {(34)}^{2} \\ \\ {( {x}^{2} )}^{2} + {( \frac{1}{ {x}^{2} }) }^{2} + 2 \times {x}^{2} \times \frac{1}{ {x}^{2} } = 1156 \\ \\ {x}^{4} + \frac{1}{ {x}^{4} } + 2 = 1156 \\ \\ {x}^{4} + \frac{1}{ {x}^{4} } = 1156 - 2 \\ \\ {x}^{4} + \frac{1}{ {x}^{4} } = 1154$

Hope this helps!!

If you have any doubt regarding to my answer, feel free to ask in the comment section or inbox me if needed.

@Mahak24

Thanks...
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