Solve following equations by factoring.
a) x2 - x = 12
b) y 2 + 12 y + 36 = 0
Solution
a) x2 - x = 12
First to solve it get everything on side of the equation and then factor.
x2 - x = 12
( x - 4) ( x + 3) = 0
Now we've got a product of two terms which is equal to zero. It means that at least one of the following must be true.
x - 4 = 0 OR x + 3 = 0
x = 4 OR x = -3
Note that each of these is linear equation i.e easy enough to solve. Now we have two solutions to the equation,
x = 4 and
x = -3 .
As through linear equations we can always check our solutions through plugging the solution back into the equation. We will check x = -3 and leave the other to you to check.
12 = 12 OK
b) y 2 + 12 y + 36 = 0
In this case already we have zero on one side & thus we don't have to do any manipulation to the equation all that we have to do is factor. Also, don't get excited regarding the fact that now we have y's in the equation. We won't always be dealing along with x's so don't expect to always see them.
So, let's factor this equation.
y 2 + 12 y + 36 = 0
(y + 6)2 = 0
(y + 6) ( y +6) = 0
In this we've got a perfect square. We broke up the square to indicate that we actually do have an application of the zero factor property. Though, we usually don't do that. Usually we will go straight to the answer from the squared part.
In this case solution to the equation is,
y = -6
We have a single value here only as opposed to the two solutions we've been getting to this point. We will frequently call this solution a double root or say that it contain multiplicity of 2 since it came from a term that was squared.