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June 10th, 2018, 04:11 AM  #1 
Newbie Joined: Jun 2018 From: Brazil Posts: 3 Thanks: 0  Problem while solving
The statement was "Knowing that a + 1 / a = b determines, as a function of b:" a) (a² + 1/a²). So I solved that doing this: (a² + 1/a²) = [(a+1/a)²  2] = b²  2 And that is clearly right. b²  2 is the right answer. But then, while solving this: c) (a^4 + 1/a^4). I tried to think the same way, so I did this: (a^4 + 1/a^4) = (a² + 1/a²)(a² + 1/a²) = [(a+1/a)²  2] [(a+1/a)²  2] = (b²  2) (b²  2) and I got b^4  4b² + 4 And that's not the right answer. Here it says the right answer is b^4  4b² + 2 Why was my way of solving it wrong? 
June 10th, 2018, 05:49 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 19,286 Thanks: 1681 
You correctly found that a² + 1/a² = (a + 1/a)²  2 = b²  2. By replacing a with a², a^4 + 1/a^4 = (a² + 1/a²)²  2 = (b²  2)²  2 = b^4  4b² + 2. 
June 10th, 2018, 06:58 AM  #3 
Newbie Joined: Jun 2018 From: Brazil Posts: 3 Thanks: 0  
June 10th, 2018, 01:02 PM  #4 
Global Moderator Joined: Dec 2006 Posts: 19,286 Thanks: 1681 
For any x and y, (x + y)² ≡ x² + y² + 2xy, so x² + y² ≡ (x + y)²  2xy. By replacing x with a and y with 1/a, a² + 1/a² ≡ (a + 1/a)²  2 (for nonzero a). By replacing x with a² and y with 1/a², a$^4$ + 1/a$^4$ ≡ (a² + 1/a²)²  2 (for nonzero a). Note that 2a(1/a) = 2 and 2a²(1/a²) = 2. 
June 10th, 2018, 01:15 PM  #5 
Global Moderator Joined: May 2007 Posts: 6,556 Thanks: 600 
Your question was solve for $a$ given $a+\frac{1}{a}=b$. You can get a simple quadratic $a^2ab+1=0$ and solve to get $a=\frac{b\pm \sqrt{b^24}}{2}$.


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