We will be using it in several sections in later

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Unformatted text preview: =8 Step 4 : Now, at this point notice that we can use the square root property on this equation. That was the purpose of the first three steps. Doing this will give us the solution to the equation. x-3 = ± 8 Þ x = 3± 8 And that is the process. Let’s do the remaining parts now. [Return to Problems] (b) 2 x 2 + 6 x + 7 = 0 We will not explicitly put in the steps this time nor will we put in a lot of explanation for this equation. This that being said, notice that we will have to do the first step this time. We don’t have a coefficient of one on the x2 term and so we will need to divide the equation by that first. Here is the work for this equation. x 2 + 3x + 7 =0 2 2 æ3ö 9 ç ÷= 4 è2ø 7 x + 3x = 2 9 79 x 2 + 3x + = - + 4 24 2 2 3ö 5 æ çx+ ÷ = 2ø 4 è x+ 3 5 =± 2 4 Þ 3 5 x=- ± i 22 Don’t forget to convert square roots of negative numbers to complex numbers! [Return to Problems] (c) 3 x - 2 x - 1 = 0 Again, we won’t put a lot of explanation for this problem. 2 2 1 x2 - x - = 0 3 3 2 1 x2 - x = 3 3 At this point we should be careful about computing the number for comple...
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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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