Review1-2011-Solns

# Review1-2011-Solns - MATH 127: Exam 1 Review Solutions...

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MATH 127: Exam 1 Review Solutions Monday, February 7, 2011 1. x 2 + y 2 + z 2 2 x + 4 y + 6 z - 14 x 2 - 2 x + y 2 - 4 y + z 2 - 6 z ≥ - 14 ( x - 1) 2 + ( y - 2) 2 + ( z - 3) 2 - 14 ≥ - 14 ( x - 1) 2 + ( y - 2) 2 + ( z - 3) 2 0 Since the square of a real number is always nonnegative, the last line is always true. Thus x 2 + y 2 + z 2 2 x + 4 y + 6 z - 14 for all x,y,z R . 2. x 2 - y 2 = ( x - y )( x + y ). Since x,y N , x - y < x + y and x + y > 0. Since 3 is prime, x + y > 0 and x - y < x + y , ( x - y )( x + y ) = 3 iﬀ ( x - y ) = 1 and ( x + y ) = 3. Solving these two equations we get x = 2 and y = 1. 3. Similar to the last analysis we have that x - y < x + y , x + y N . If ( x - y )( x + y ) = 15 then x - y N since x + y N . 15 = 1 · 15 = 3 · 5, and these are the only 2 ways to decompose 15, since it is a product of 2 primes. Thus either x - y = 1 and x + y = 15 or x - y = 3 and x + y = 5. Thus the only 2 solutions are (8 , 7) and (4 , 1). 4.

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## This note was uploaded on 09/08/2011 for the course MATH 21-127 taught by Professor Gheorghiciuc during the Spring '07 term at Carnegie Mellon.

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Review1-2011-Solns - MATH 127: Exam 1 Review Solutions...

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