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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: 20,926 Thanks: 2205 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, 06:58 AM   #3
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 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, 01:02 PM #4 Global Moderator   Joined: Dec 2006 Posts: 20,926 Thanks: 2205 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, 01:15 PM #5 Global Moderator   Joined: May 2007 Posts: 6,806 Thanks: 716 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 $a=\frac{b\pm \sqrt{b^2-4}}{2}$. Tags problem, solving Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post redsaints2010 Probability and Statistics 6 April 2nd, 2017 03:41 PM MsHalifax Elementary Math 9 June 3rd, 2016 06:13 PM tayyab_zarif Elementary Math 1 December 10th, 2012 03:20 PM tayyab_zarif Elementary Math 0 December 4th, 2012 02:17 AM beefeater267 Real Analysis 3 August 30th, 2007 07:25 AM

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