SOLUTION: Please help me solve this equation: If x = 3 + 2sqrt(2), what is the value of x^4 + (1/x^4) ?

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Question 632628: Please help me solve this equation: If x = 3 + 2sqrt(2), what is the value of
x^4 + (1/x^4) ?

Answer by Edwin McCravy(20055) About Me  (Show Source):
You can put this solution on YOUR website!
To do this requires thinking of the fact that  

(A+1%2FA%29)² = (A+1%2FA%29)(A+1%2FA%29) = A² + 2A·(1%2FA + 1%2FA%5E2 = A² + 2 + 1%2FA%5E2 = A² + 1%2FA%5E2 + 2.

Therefore

A² + 1%2FA%5E2 + 2 = (A+1%2FA%29)²

or

A² + 1%2FA%5E2 = (A+1%2FA%29)² - 2
 
The theorem which the above proves:

The sum of a square and its reciprocal equals to the square of the
sum of their two square roots decreased by 2

x4 + 1%2Fx%5E4 

That's the sum of a square and its reciprocal, so using the 
theorem above it becomes

(x² + 1%2Fx%5E2)² - 2

What's inside the parentheses is also the sum of a square
and its reciprocal, so using the theorem above it becomes

[(x + 1%2Fx)² - 2]² - 2

Now we need to find 

                1%2Fx by substituting 3 + 2sqrt%282%29 for x  
                
                1%2Fx = 1%2F%283%2B2sqrt%282%29%29 = 1%2F%283%2B2sqrt%282%29%29 =
                1%2F%283%2B2sqrt%282%29%29%283-2sqrt%282%29%29%2F%283-2sqrt%282%29%29 = %283-2sqrt%282%29%29%2F%289-4%2A2%29 = %283-2sqrt%282%29%29%2F%289-8%29 =
                 %283-2sqrt%282%29%29%2F1 = 3 - 2sqrt%282%29

Therefore 

[(x + 1%2Fx)² - 2]² - 2

becomes

[(3 + 2sqrt%282%29 + 3 - 2sqrt%282%29)² - 2]² - 2

[(6)² - 2]² - 2

[36 - 2]² - 2

34² - 2

1156 - 2

1154

Edwin