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June 10th, 2018, 05:11 AM   #1
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Post 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?
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June 10th, 2018, 06:49 AM   #2
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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.
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June 10th, 2018, 07:58 AM   #3
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Quote:
Originally Posted by skipjack View Post
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?
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June 10th, 2018, 02:02 PM   #4
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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.
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June 10th, 2018, 02:15 PM   #5
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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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