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Unformatted text preview: MATH 135 Algebra, Solutions to Assignment 1 1: (a) Find all real numbers x such that √ x 1 + x = 1 3 √ x . Solution: By cross multiplying and then using the quadratic formula, we have √ x 1 + x = 1 3 √ x ⇐⇒ √ x (3 √ x ) = 1 + x ⇐⇒ 3 √ x x = 1 + x ⇐⇒ 2 x 3 √ x + 1 = 0 ⇐⇒ √ x = 3 ± √ 9 8 4 ⇐⇒ √ x = 1 or 1 2 ⇐⇒ x = 1 or 1 4 . (b) Solve y = x + 1 x with 0 < x ≤ 1 for x in terms of y . Solution: Multiply both sides of the equation by x then use the quadratic formula to get y = x + 1 x ⇐⇒ xy = x 2 + 1 ⇐⇒ x 2 xy + 1 = 0 ⇐⇒ x = y ± p y 2 4 2 . Note that 0 < x ≤ 1 = ⇒ 1 x ≥ 1 ≥ x = ⇒ y = x + 1 x ≥ 2 x = ⇒ x ≤ y 2 , so we must use the negative sign. Thus x = y p y 2 4 2 . 2: (a) Find all ordered pairs of integers ( x,y ) such that xy = 6 + 2 x . Solution: We have xy = 6 + 2 x ⇐⇒ xy 2 x = 6 ⇐⇒ x ( y 2) = 6. So x must be a factor of 6, that is x = ± 1 , ± 2 , ± 3 , ± 6, and we have y = 6 x + 2. Thus ( x,y ) = ( 6 , 1) , ( 3 , 0) , ( 2 , 1) , ( 1 , 4) , (1 , 8) , (2 , 5) , (3 , 4) or (6 , 3) . Note that these are the 6 points with integer coordinates which lie on the hyperbola y = 6 x + 2. (b) Find all ordered pairs of integers ( x,y ) such that x 2 + y 2 = 4 x + 2 y . Solution: Complete the square to get x 2 + y 2 = 4 x +2 y ⇐⇒ x 2 4 x + y 2 2 y = 0 ⇐⇒ ( x 2) 2 4+( y 1) 2 1 = 0 ⇐⇒ ( x 2) 2 +( y 1) 2 = 5 ....
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This note was uploaded on 07/25/2011 for the course MATH 135 taught by Professor Andrewchilds during the Spring '08 term at Waterloo.
 Spring '08
 ANDREWCHILDS
 Algebra, Quadratic Formula, Real Numbers

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