Therefore we will work only with square roots in this

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Unformatted text preview: 22 -4=0 So, if we use the substitution, © 2007 Paul Dawkins 113 College Algebra u 2 = (t 2 ) = t 4 2 u = t2 the equation becomes, u2 - 4 = 0 and so it is reducible to quadratic in form. Now, we can solve this using the square root property. Doing that gives, u = ± 4 = ±2 Now, going back to t’s gives us, u=2 Þ t2 = 2 Þ t=± 2 u = -2 Þ t 2 = -2 Þ t = ± -2 = ± 2 i In this case we get four solutions and two of them are complex solutions. Getting complex solutions out of these are actually more common that this set of examples might suggest. The problem is that to get some of the complex solutions requires knowledge that we haven’t (and won’t) cover in this course. So, they don’t show up all that often. [Return to Problems] All of the examples to this point gave quadratic equations that were factorable or in the case of the last part of the previous example was an equation that we could use the square root property on. That is need not always be the case however. It is more than possible that we would need the quadratic formula to do some of these. We should do an example of one of these just to make the point. Example 3 Solve 2 x...
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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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