Unformatted text preview: y + 36 = 0 ( y + 6) = 0
( y + 6 )( y + 6 ) = 0
2 In this case we’ve got a perfect square. We broke up the square to denote that we really do have
an application of the zero factor property. However, we usually don’t do that. We usually will
go straight to the answer from the squared part.
The solution to the equation in this case is, y = 6 We only have a single value here as opposed to the two solutions we’ve been getting to this point.
We will often call this solution a double root or say that it has multiplicity of 2 because it came
from a term that was squared.
[Return to Problems] (d) 4m 2  1 = 0
As always let’s first factor the equation. 4m 2  1 = 0 ( 2m  1)( 2m + 1) = 0
Now apply the zero factor property. The zero factor property tells us that, 2m  1 = 0
2m = 1
1
m=
2 2m + 1 = 0
2 m = 1
1
m=2 OR
OR
OR Again, we will typically solve these in our head, but we needed to do at least one in complete
detail. So we have two solutions to the equation. m= 1
2 AND m= 1
2
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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.
 Spring '12
 MrVinh

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