University of Michigan
Runs his own tutoring company
Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!
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University of Michigan
Runs his own tutoring company
Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!
Solving a quadratic by completing the square. For this problem what we’re looking at is a quadratic where we’re going to solve by completing the square but with the added problem of having a coefficient on our leading term or on our x² term.
Our first step is still going to be the same as before which is to get our xs by itself which means we have to bring the 30 to the other side. So we have 3x² minus 12x is equal to 30. In order to complete the square our leading coefficient has to be a 1 and there’s two ways of dealing with this problem.
The first thing we could do is divide it by 3 which would get rid of that coefficient make this coefficient one. The other thing we could do is factor that 3 out from all these x terms. I’m going to do the second just because later in this course we’re going to have to do some more completing the square which we’re not going to be able to divide by whatever that coefficient is so it’s really good practice right now to get used to the process that we’re going to have to use later.
What we’re going to do is just factor out this 3 leaving us with x² minus 4x, I’m going to leave a little space for the constant term that we’re going to have to add, that’s going to be equal to 30. We now want to express this middle term as a perfect square. The way we do that is we know we have a x the middle term is negative so it has to be a subtraction, and then we take this middle term, 4 divide by 2 so that’s what goes over here and then we have to add this term, squared inside the previous parentheses.
You can always check if we FOIL this, I’m just going to leave a little squared there. If we FOIL this out we would actually get this here.
One of the problems that people forget about is we have added 4 inside the parentheses. So we’ve added 4 in here, but what have we really added to this side as a whole? We have this 3 out in the front so what we’ve really added is this, 3 times 4. We’ve actually added 12 to this side. Make sure you remember to distribute that through. So in order to keep the equation balanced we have to add 12 to the other side of the equation as well.
So this is then equal to 42 and I need to bring that down this coefficient of 3 that was on the outside. We now have a equation which we can solve. First thing we want to do is divide by that 3, okay we had to divide by it at some point but it’s a little bit easier to do afterwards than it was before. We have x minus 2² is equal to 42 divide by 12 is I believe 14. Take the square root of both sides in order to solve for x, we end up with x minus 2 is equal to the plus or minus the square root of 14. Remember whenever you use the square root as a tool, include your plus or minus, add two to both sides, finishing up as x is equal to 2 plus or minus the square root of 14.
So completing the square to solve a quadratic equation when we have a coefficient out in front, is pretty much the same exact thing, just factor out that coefficient at the beginning and when you’re adding in your number to make it a perfect square, always, always, always remember to distribute through to that coefficient to make sure you keep it balanced when you add it to the other side.
Unit
Quadratic Equations and Inequalities