The most common mistake is to decide that the first

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Unformatted text preview: , the one is a solution. Now let’s check x = 5 . OK 2 ( 5) - 1 - 5 - 4 = 2 ? ? 9 - 1=2 3 -1 = 2 OK So, they are both solutions to the original equation. [Return to Problems] © 2007 Paul Dawkins 120 http://tutorial.math.lamar.edu/terms.aspx College Algebra (b) t + 7 + 2 = 3 - t In this case we’ve already got a square root on one side by itself so we can go straight to squaring both sides. ( t+7 +2 ) =( 2 3-t ) 2 t + 7 + 4 t + 7 + 4 = 3-t t + 11 + 4 t + 7 = 3 - t Next, get the remaining square root back on one side by itself and square both sides again. 4 t + 7 = -8 - 2t ( 4 t+7 ) 2 = ( -8 - 2t ) 2 16 ( t + 7 ) = 64 + 32t + 4t 2 16t + 112 = 64 + 32t + 4t 2 0 = 4t 2 + 16t - 48 0 = 4 ( t 2 + 4t - 12 ) 0 = 4 ( t + 6 )( t - 2 ) Þ t = -6, t = 2 Now check both possible solutions starting with t = 2 . ? 2+ 7 + 2= 3- 2 ? 9 + 2= 1 3+ 2 ¹1 NOT OK So, that wasn’t a solution. Now let’s check t = -6 . -6 + 7 + 2 = 3 - ( -6 ) ? ? 1+ 2= 9 1+ 2 = 3 OK It looks like in this case we’ve got a single solut...
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This note was uploaded on 06/06/2012 for the course ICT 4 taught by Professor Mrvinh during the Spring '12 term at Hanoi University of Technology.

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