Homework 1 Solution

Homework 1 Solution - MATH 74 HOMEWORK 1 DUE WEDNESDAY 9/5...

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MATH 74 HOMEWORK 1: DUE WEDNESDAY 9/5 1(a). Find all real numbers x satisfying x 4 - 6 x 2 + 7 = 0. If x satisfies the given equation, then the real number t = x 2 satisfies t 2 - 6 t +7 = 0. Applying the proposition and simplifying a little, we see that either t = 3 + 2 or t = 3 - 2; we conclude that either x 2 = 3 + 2 or x 2 = 3 - 2. Applying the proposition again in either of these cases, or just taking square roots, we see that x must one of the numbers p 3 + 2, - p 3 + 2, p 3 - 2, - p 3 - 2. Conversely, if x happens to be any one of those numbers listed, we see that x 2 is one of the numbers 3 ± 2, so that (by the proposition again) x 4 - 6 x 2 + 7 = 0. Conclusion: the set of real numbers x satisfying the equation is precisely the set { p 3 + 2 , - p 3 + 2 , p 3 - 2 , - p 3 - 2 } . 1(b). Find all real numbers x satisfying x 4 - 2 x 2 - 1 = 0. If x is a real number satisfying the given equation, then t = x 2 satisfies t 2 - 2 t - 1 = 0, and hence (by the proposition and some algebra) is one of the numbers
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This note was uploaded on 02/10/2010 for the course MATH 74 taught by Professor Courtney during the Fall '07 term at Berkeley.

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Homework 1 Solution - MATH 74 HOMEWORK 1 DUE WEDNESDAY 9/5...

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