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Ch7_EquIneq - MA100 Fall 2010 PART 1 Precalculus[7...

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MA100, Fall 2010, PART 1: Precalculus [7] Equations and Inequalities Note : In this Chapter, unless otherwise stated, all variables (all x, y, , z etc.) are considered real numbers. 1. Equations Solving an equation does not always lead an answer that consists in an unique object (number), but that sometimes the answer is that there is no solution , or there are two or more solutions (sometimes there are infinitely many solutions). (Think for instance of the quadratic equation ax 2 + bx + c = 0 , a, b, c R , a = 0, which can have two distinct real solutions, one real solution or no real solutions.) Up to know you should know how to: 1. solve linear equations, i.e. equations of the form ax + b = 0 , where a, b are given real numbers (see online notes LinearIneq ); 2. solve quadratic equations, i.e. equations of the form ax 2 + bx + c = 0 , where a, b, c are given real numbers with a = 0 (see online notes Algebra). The objective of this chapter is to discuss and solve (if possible) equations that take the form: (A) math. expression A = 0 , and (B) math. expression A math. expression B = 0 . (A) Solving equations of the type math. expression A = 0 The first thing to do is to factor math. expression A as mush as possible. After factoring, it is important to remember that: A product is zero if at least one its factors is zero! Example: Solve x ( x - 2)(2 x + 1) = 0 . Solution: x ( x - 2)(2 x + 1) = 0 = x = 0 or ( x - 2) = 0 or (2 x + 1) = 0 = x = 0 or x = 2 or 2 x = - 1 = x = 0 or x = 2 or x = - 1 2 In set notation, the answer is “The solution set is 0 , 2 , - 1 2 ”. The answer can also be written as x 0 , 2 , - 1 2 1
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Example: Solve x 3 - 3 x + 2 = 0 . Solution: First we factor x 3 - 3 x + 2 = 0. Using either the algorithm for finding factors of degree 1 (see online notes Algebra), or directly, we have x 3 - 3 x + 2 = ( x - 1)( x 2 + x - 2) . ( x - 1)( x 2 + x - 2) = 0 = ( x - 1) = 0 or ( x 2 + x - 2) = 0 = x = 1 or x 2 + x - 2 . To solve x 2 + x - 2 = 0 we use the quadratic formula. We have a = 1 , b = 1 and c = - 2 . Δ = b 2 - 4 ac = 1 2 - 4 · 1 · ( - 2) = 1 + 8 = 9 and so x 1 , 2 = - 1 ± 9 2 = x 1 , 2 = - 1 ± 3 2 = x 1 = - 1+3 2 = x 1 = 2 2 = x 1 = 1 or x 2 = - 1 - 3 2 = x 2 = - 4 2 = x 2 = - 2 So overall x = 1 or x = 1 or x = - 2 . or, since x = 1 is repeated, just x = 1 or x = - 2 . The answer in set notation is The solution set = { 1 , - 2 } , or x ∈ { 1 , - 2 } . Note : You are free to write the answer in any form you wish as long as your answer is correct and it is written in a correct form ! Example: Solve ( x 2 - x - 1)( x 2 - 4) = 0 . Solution: ( x 2 - x - 1)( x 2 - 4) = 0 = ( x 2 - x - 1)( x - 2)( x + 2) = 0 = x 2 - x - 1 = 0 or x - 2 = 0 or x + 2 = 0 . = x 2 - x - 1 = 0 or x = 2 or x = - 2 . 2
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We need to solve separately x 2 - x - 1 = 0 . Using the (quadratic) formula for ax 2 + bx + c = 0 we have a = 1 , b = ( - 1) and c = ( - 1) . Further, Δ = b 2 - 4 ac = ( - 1) 2 - 4 × 1 × ( - 1) = 1 + 4 = 5 and so x 1 , 2 = - b ± Δ 2 a = - ( - 1) ± 5 2 = 1 ± 5 2 = x 1 = 1+ 5 2 or x 2 = 1 - 5 2 Returning to our initial equation, we have ( x 2 - x - 1)( x 2 - 4) = 0 = . . .
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