Misc. Study Materials - Algebra and Precalculus Review...

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Page 1 of 3 Algebra and Precalculus Review Selected solutions Aug 21, 2011 A “simplified” expression 1. 48 2 27 4 ± = x 48 2 27 4 ± = via the quadratic formula 2. 2 2 2 ) 5 ( 5 2 3 x x x x + + + = 0 ) 5 ( ) 1 )( 5 3 ( 2 2 > + + + x x x x for x in ) , 0 ( ) 0 , 1 ( 3 5 3. 2 3 8 2 1 x x 0 2 16 2 3 < = x x for x in ) 16 , 0 ( 4. From the simplified version, 2 3 2 ) 4 3 ( 3 8 3 ) ( x x x x f + = , one can readily identify where f is undefined (the zeros of the denominator). Thus the domain is ) , ( ) , 0 ( ) 0 , ( 3 4 3 4 −∞ . Exponents 1. (a) 4 2 3 z (b) 4 1 ba (c) ) 2 ( ) 2 ( 3 2 + x x (d) 2 22 (e) ) 1 3 ( 8 + + x x x (f) 25 2 ) 5 2 ( x x x Distrubutive property 1. Multiply and collect like terms: (a) (b) 3 5 2 2 + δδ 2 2 9 49 εε + Δ Δ (c) (d) 15 4 24 2 + + x x 3 4 3 2 5 5 x x = 3 4 2 ) 1 ( 5 x x 2. Factor the expression (a) (b) 2 ) 1 2 ( x ) 2 )( 1 )( 1 ( 12 + x x x (c) ) 2 )( 4 ( ) 1 2 ( 7 2 x x x (d) 4 ) 2 1 ( ) 7 )( 2 ( 2 x x x + + (e) 1 3 1 2 2 + x x Fractions 1. (a) ) 2 )( 1 ( ) 1 )( 3 2 ( + x x x x (b) ) 2 ( ) 1 ( ) 7 22 )( 7 22 ( 2 + + x x x x (c) ) 4 ( 4 ) 3 13 )( 2 ( + x x x x (d) 2 3 2 16 x x 2. (a) -8/5 (b) 7 3 1 x (c) ) 1 2 ( 4 3 2 + x x Solve the Equation or Inequality 1. Solve the Equation (c) 3 1 , 2 , 0 = = = x x x (d) 3 2 ± = x (e) 6 7 ± = t (f) 20 = x 2. Solve the inequality (a) [ (b) (c) ] 5 , 2 ) , 14 ( ) 6 , ( −∞ ) 2 , 7 ( (d) ) , 2 ( ] 2 3 ,
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This note was uploaded on 10/15/2011 for the course MISC. 101 taught by Professor Multiple during the Fall '11 term at University of Texas at Dallas, Richardson.

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Misc. Study Materials - Algebra and Precalculus Review...

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