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Unformatted text preview: ions by factoring.
(a) x 2 - x = 12 [Solution]
(b) x 2 + 40 = -14 x [Solution]
(c) y 2 + 12 y + 36 = 0 [Solution]
(d) 4m 2 - 1 = 0 [Solution]
(e) 3 x 2 = 2 x + 8 [Solution]
(f) 10 z 2 + 19 z + 6 = 0 [Solution]
(g) 5 x 2 = 2 x [Solution]
Now, as noted earlier, we won’t be putting any detail into the factoring process, so make sure that
you can do the factoring here.
(a) x 2 - x = 12
First, get everything on side of the equation and then factor. x 2 - x - 12 = 0 ( x - 4 )( x + 3) = 0
Now at this point we’ve got a product of two terms that is equal to zero. This means that at least
one of the following must be true. x-4=0
x=4 x+3= 0
x = -3 OR
OR Note that each of these is a linear equation that is easy enough to solve. What this tell us is that
we have two solutions to the equation, x = 4 and x = -3 . As with linear equations we can
always check our solutions by plugging the solution back into the equation. We will check
x = -3 and leave the other to you to check. ( -3) - ( -3) =12
2 ? ? 9 + 3 = 12
12 = 12 OK So, this was in fact a solution.
[Return to Problems] (b) x 2 + 40 = -14 x
As with th...
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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