Getting complex solutions out of these are actually

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Unformatted text preview: [Solution] (d) t 4 - 4 = 0 [Solution] Solution 2 1 (a) x 3 - 2 x 3 - 15 = 0 Okay, in this case we can see that, 2 æ1ö = 2ç ÷ 3 è3ø and so one of the exponents is twice the other so it looks like we’ve got an equation that is reducible to quadratic in form. The substitution will then be, u=x 2 2 æ 1ö u = ç x3 ÷ = x3 èø 1 3 2 Substituting this into the equation gives, u 2 - 2u - 15 = 0 ( u - 5 )( u + 3) = 0 Þ u = -3, u = 5 Now that we’ve gotten the solutions for u we can find values of x. 1 3 u = -3 : x = -3 u =5: x3 = 5 1 Þ x = ( -3) = -27 Þ x = ( 5 ) = 125 3 3 So, we have two solutions here x = -27 and x = 125 . [Return to Problems] (b) y -6 - 9 y -3 + 8 = 0 For this part notice that, © 2007 Paul Dawkins 112 http://tutorial.math.lamar.edu/terms.aspx College Algebra -6 = 2 ( -3) and so we do have an equation that is reducible to quadratic form. The substitution is, u 2 = ( y -3 ) = y -6 2 u = y -3 The equation becomes, u 2 - 9u + 8 = 0 ( u - 8)( u - 1) = 0 u = 1, u = 8 Now, going back to y’s is going to take a little more work here, but shouldn’t be too bad. u = 1: u = 8: 1 =1 y3 Þ y -...
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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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