CHAPTER 03.01: QUADRATIC EQUATIONS: Derivation of the Quadratic Formula

In this segment we're going to derive the quadratic solution, so, for the quadratic formula . . . quadratic equation.  So we have a quadratic equation somebody gives this to us, a general one, and we want to derive the solution.  So you've already seen that if you have a x squared plus b x plus c equal to 0, we already know that the solution looks like this, this quite popular formula.  So that is the solution to the quadratic equation, and what we're going to do in this segment, we're going to derive this particular formula, that how does it turn out to be that the solution is as follows.  Now, when you have a x squared plus b x plus c equal to 0, that's the quadratic equation, we're going to assume that a is not equal to 0, because if a is equal to 0, then we know that we have b x plus c equal to 0, that's what the equation will turn out to be, a is equal to 0, this becomes 0, so you have b x plus c equal to 0, so you get x equal to -c divided by b, that's what you're going to get for the solution.  So it is safe to assume that we are going to solve a quadratic equation which only has . . . which has a not equal to 0, because we know for a equal to 0 the solution exists, which looks like that.  So let's . . . in assuming that a is not equal to 0, we have a x squared plus b x plus c equal to 0, and what I'm going to do is I'm going to divide by a throughout, so I get this. So this is . . . we obtained this by divide by a, that's what I get, so if I divide by a both sides, so since I'm already assumed that a is not equal to 0, so let me write this down again here, a not equal to 0.  So I get . . . I get this particular formula.  Now what I'm going to do is I'm going to add and subtract a quantity here, so I'm going to add b squared by 4 a squared here, I'm going to subtract b squared by 4 a squared here, and then plus b by a x . . . sorry, plus c by a equal to 0.  So what I'm doing is I am taking this quantity here, adding it, and then I'm subtracting it immediately, and that gives me 0, so that reproduces the equation which I had before there.  But what I can do is I can bundle this together. I can bundle the first three terms of this term together, and they will turn out to be x plus b divided by 2 a, whole squared, so that's from your formula of a plus b, whole squared, that's what I get that from, and then I have this quantity of minus b squared by 4 a squared plus c by a equal to 0.  Having said that, what I'm going to do is I'm going to take the last two terms of the left-hand side to the right-hand side, so I'm going to take this to the right-hand side, I'll get b squared by 4 a squared plus . . . minus c by a . . . minus c by a, that's what I get there.  And I'm going to simplify this a little bit, I'm going to get 4 a squared here, and I'll do b squared minus 4 a c here, so that's what it turns out to be. And now what I'm going to do is I'm going to take the square root of both sides, I'm going to take the square root of both sides here, and that's where I get the plus and the minus part of it there, because when I take the square root of both sides, this can . . . this can be plus or minus. So, let me write that down.  So I'll get x plus b divided by 2 a, so that's taken by doing the square root of both sides, is equal to square root of b squared minus 4 a c divided by 4 a squared, that's what I'm going to get from there.  Then what I can do is I could write this as b squared minus 4 a c, divided by 2 a, because the square root of 4 a squared is 2 a, so this should be plus and minus, because both are possibilities when you take the square root of both sides, so plus minus, I'm going to put plus minus there.  And then I get x is equal to -b divided by 2 a, by taking this b divided by 2 a to the right-hand side, plus minus square root of b squared minus 4 a c, divided by 2 a, there.  And then I'm going to write x is equal to, I'm going to take 2 a common in the denominator, so then I'll write -b plus minus b squared minus 4 a c. So that's the kind of form which you see as the solution for a quadratic equation, which is the form a x squared plus b x plus c equal to 0, is solution of this.  So that's how we derive the solution for a quadratic equation, the two roots of the quadratic equation, a x squared plus b x plus c equal to 0.  And that's the end of this segment.