Alg_Complete

# So we have two solutions to the equation m 1 2 and m

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Unformatted text preview: e first one we first get everything on side of the equal sign and then factor. x 2 + 40 + 14 x = 0 ( x + 4 )( x + 10 ) = 0 Now, we once again have a product of two terms that equals zero so we know that one or both of them have to be zero. So, technically we need to set each one equal to zero and solve. However, this is usually easy enough to do in our heads and so from now on we will be doing this solving in our head. © 2007 Paul Dawkins 86 http://tutorial.math.lamar.edu/terms.aspx College Algebra The solutions to this equation are, x = -4 x = -10 AND To save space we won’t be checking any more of the solutions here, but you should do so to make sure we didn’t make any mistakes. [Return to Problems] (c) y 2 + 12 y + 36 = 0 In this case we already have zero on one side and so we don’t need to do any manipulation to the equation all that we need to do is factor. Also, don’t get excited about the fact that we now have y’s in the equation. We won’t always be dealing with x’s so don’t expect to always see them. So, let’s factor this equation. y 2 + 12...
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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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