Solving Quadratic Equations by Completing the Square

All right, so we are going to continue with solving quadratics and today we're going to look at a method called completing the square. So Lake stated before all quadratics are solvable. But not everything is factorable, so we have to have other methods for solving quadratics. So we're going to start off with this example. X squared plus two X -7 equals zero. And we should look for factoring first, because it is an easy method. In this case, we would be looking for what multiplies to get negative 7 adds to get positive two. And that's not possible. There's nothing that multiplies to get negative 7 ads to get positive two. So this can not be solved by factoring. So when we talk about completing the square, we have a list of steps to complete when doing this process. So we're dealing with a standard form of a quadratic. And the first step is to collect variable terms on one side of the equation and constants on the other. So unlike factoring, we do not want this set equal to zero. We need this set equal to the constant. So we're going to move that 7 to the other side. Sometimes this step will already be done for you, and it'll be set equal to the constant. Other times you'll have to move terms around. But you always want the X squared and X on one side, the constant, on the other. The next step is an if necessary step, as needed, divide both sides by a to make the coefficient of the X squared term one. We always want the leading coefficient to be one when we're doing completing the square. In this problem, we already have a leading coefficient of one. The coefficient of X squared, we don't need to divide in this case. So we'll look at an example problem here in a few minutes where we would have to divide. The next step is the process of actually completing the square. And this is referred to as completing the square because we are going to make the left side of this equation a perfect square trinomial. To do that, we are going to add B over two squared to both sides of the equation. So in this case, we have two over two squared. Well, two over two is one, one squared is just one. So we are going to add one to both sides of this problem. By adding one to both sides, we're not changing the value any. Now, we can factor this perfect square trinomial. X squared plus two X plus one factors to X plus one times X plus one. We can add here on the right to get a. Remember, X plus one times X plus one can be rewritten as X plus one squared. And we can solve this. We know we can take the square root of both sides. So we get X plus one equals plus or minus square root of 8, well that can be simplified. And so that breaks down two square root of two. And then the last step is just to solve for the values of the variable. That means get the variable by itself by subtracting. So the solution to this quadratic X squared plus two X -7 equals zero. Is negative one plus or minus two square root of two. Always remember to simplify the radicals. And don't forget the plus or minus when you take the square root sign. I highly recommend pausing the video and copying down this chart. This will be very helpful in working through problems by completing the square. So let's look at another example. Try to make wearing still see the chart. So in this case, our example, we're going to have two X squared -8 X. Equals 22. So we're solving by completing the square, the first step is to collect variable terms on one side, constants on the other. Well, that step is already taken care of. X squared and X are on the left, the constants on the right. Step two is divide by a to make the leading coefficient one. So in this case, we have to divide the entire equation by two. It is important to not forget this step. So X squared minus four X equals 11. Step three is to complete the square by adding B over two squared to both sides. So remember B over two squared. So be in this case is negative four. Negative four over two is negative two. Negative two squared is four. So we're going to add four to both sides. Of the equation. By adding four to both sides, we have not changed the value. We've now created a perfect square trinomial, so we factor that. X and X remember we're looking for what multiplies to get four, adds to get negative two. Negative four, so it is negative two. I'm going to go ahead and add. This product can be rewritten as X minus two. Squared. And then we can work on solving for X we can take the square root of both sides of the equation. So we have X minus two equals plus or -15 square root of 15. And really, we don't actually need these parentheses around, let's see if I can erase this. Let's try that again. When we square rooted that, we didn't need the parentheses anymore. And then the last step would be to add the two over. And so we get X equals two plus or minus the square root of 15. Again, it's really important on this example that we divide by the leading coefficient. You always need to make sure you do that step. All right. Let's look at another one. So in this, we've got. X squared equals 27 -6 X. So again, the first step is going to be to get the variables on one side. So I'm going to add the 6 X to the left. And we'll leave the constant here. Now, we need to complete the square by adding B over two squared to both sides. And a lot of times, in other videos, you may see that they kind of do this little fill in the blank concept. That might help in figuring out what needs to be added to both sides and keeping the correct format. So in this case, we have 6 over two. Squared or three squared, so we're going to add 9 to both sides. Factor the trinomial. So in this case, you're looking for what multiplies to get 9, adds to get 6. So three and three. Add to get 36. Rewrite the product as a binomial squared. And then we can take the square root of both sides. Remember, anytime you take the square root of both sides, it's going to have plus minus involved. And in this case, 36 is a perfect square so that would just be 6. Subtract the three. Since these are both like terms since we no longer have a radical involved. We can actually break this into two problems, X equals negative three plus 6. And X equals negative three -6. To get our actual solutions of three and negative 9. Any time this scenario happens where you no longer have a square root, you need to break it into the two equations or two problems and simplify to get your two answers. All right, so make sure you write down the chart and we will look at more examples of completing the square in our next Google Meet.