My Math Forum Problem while solving

 Algebra Pre-Algebra and Basic Algebra Math Forum

 June 10th, 2018, 05:11 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, 06:49 AM #2 Global Moderator   Joined: Dec 2006 Posts: 19,967 Thanks: 1850 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. Thanks from niklaus
June 10th, 2018, 07:58 AM   #3
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Quote:
 Originally Posted by skipjack 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.
Ok. But why the -2 after the (a²+1/a²)² ? Where does it come from?

 June 10th, 2018, 02:02 PM #4 Global Moderator   Joined: Dec 2006 Posts: 19,967 Thanks: 1850 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 non-zero a). By replacing x with a² and y with 1/a², a$^4$ + 1/a$^4$ ≡ (a² + 1/a²)² - 2 (for non-zero a). Note that 2a(1/a) = 2 and 2a²(1/a²) = 2.
 June 10th, 2018, 02:15 PM #5 Global Moderator   Joined: May 2007 Posts: 6,641 Thanks: 625 Your question was solve for $a$ given $a+\frac{1}{a}=b$. You can get a simple quadratic $a^2-ab+1=0$ and solve to get $a=\frac{b\pm \sqrt{b^2-4}}{2}$.

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