Not only that but what weve got left here is

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Unformatted text preview: aring both sides. 1 - t = 2t - 3 (1 - t ) 2 = ( 2t - 3 ) 2 1 - 2t + t 2 = 2t - 3 t 2 - 4t + 4 = 0 (t - 2) 2 =0 Þ t=2 So, we have a double root this time. Let’s check it to see if it really is a solution to the original equation. © 2007 Paul Dawkins 118 http://tutorial.math.lamar.edu/terms.aspx College Algebra 1= 2 + 2 ( 2) - 3 ? ? 1= 2 + 1 1¹ 3 So, t = 2 isn’t a solution to the original equation. Since this was the only possible solution, this means that there are no solutions to the original equation. This doesn’t happen too often, but it does happen so don’t be surprised by it when it does. [Return to Problems] (c) 5 z + 6 - 2 = z This one will work the same as the previous two. 5z + 6 = z + 2 ( 5z + 6 ) 2 = ( z + 2) 2 5z + 6 = z 2 + 4z + 4 0 = z2 - z - 2 0 = ( z - 2 )( z + 1) Þ z = -1, z = 2 Let’s check these possible solutions start with z = -1 . 5 ( -1) + 6 - 2 = - 1 ? ? 1 - 2 = -1 - 1 = -1 So, that’s was a solution. Now let’s check z = 2 . OK 5 ( 2) + 6 - 2 = 2 ? ? 16 - 2 = 2 4-2 = 2 OK This was also a solution. So, in this case we...
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