Deriving the Quadratic Formula

The quadratic formula is one of the most important formulas that you will learn in Algebra and chances are that you have probably memorized it. But where does it come from When deriving the quadratic formula, we first start with a generic quadratic formula using coefficients a, b and c and then derive the formula by completing the square.

So hopefully by now you have seen and are familiar with the quadratic formula. Negative B plus or minus the square root of B squared minus 4AC all over 2A.

You may have a song for it, you may have some different way of memorizing it but hopefully you've seen it in some context. What I want to do now is actually spend some time and show you how this is actually derived.

Okay. So you know that it's not some crazy formula that comes out of thin air. Actually what ends up happening it's found by completing the square with a general quadratic using coefficients A, B and C. So what we're going to do is complete the square going to the exact same process but instead of doing numbers we're going to have letters.

So the first thing we did when we were completing the square is to move the C over to the other side. So we'll go ahead and do that. X squared plus BX is equal to negative C. We then -- oops, why did I put a C there. It should be X. Try that again. All right. So we then factored out the coefficient on our leading term. So that means we take out the A. Take out the A. This becomes X squared. When we take out A from B, this is just going to be left with B over AX and then equals negative C.

You can always check this because when you multiply that A back through we should end up with B. Here A and A cancels leaving us with B. So from here our goal is to make this be a perfect square.

Okay. And there's something that we have to add in here which corresponds to this blank spot we had here. How we found that was by taking this middle term and dividing it by 2. That's what went right in this box. So that ends up with B divided by 2A. That term squared is what we ended up adding over here. So when we square B over 2A what we end up is B squared over 4A squared.

Remember that we have to distribute this coefficient in. This A -- we're adding B squared over 4A squared inside the parentheses, but what we're really doing is adding an A times that to that entire side. So we added the B squared over 4A squared to this side. Multiplying it by A. We have to add it to the other side. So that's plus B squared over 4A. One of those As cancel when we multiply it in.

All right. So we have A times this quantity squared is equal to negative C plus B squared over 4A. Okay.

We now want to try to get our square terms by itself. In order to do that we need to divide by A. Giving us X plus B over 2A squared is equal to negative C over A plus B squared over 4A squared. So this A got distributed to both terms. Leaving us right here. Okay.

So now we need to take the square root of both sides. The square root of a square is just going to cancel out. Let's go over here and write it up. What we end up with then is just the X plus B over 2A. And that is going to be equal to plus or minus the square root of this side. Plus or minus square root of negative CA plus B squared. Over 4A squared. All right.

Before I continue on, what I want to do is actually combine things in this square root. In order to combine things we need a common denominator. So common denominator in this case is 4A squared. So this term needs a 4A over 4A. The left side stays the same. X plus B over 2A is equal to plus or minus square root of negative 4AC over 4A squared plus B squared over 4A squared. All right.

Solving for X what we have to do then is subtract the B over 2A over. X is equal to plus or minus this giant ugly square root. And for this one I'm actually going to combine our terms. We have our common denominator. You can throw them together. Negative 4AC plus B squared over 4A squared and then we minus, make this a little bit shorter, B over 2A. All right. We're almost there.

So what we have left is this term right here. We can distribute this radical to both the numerator and the denominator. I'm running out of room over here. And lastly what we have to do is just a little bit of rearranging. We have plus or minus the square root, negative 4 AC plus B squared. We rearrange that and we end up with B squared minus 4 AC, square root of 4 A squared is just 2A. Where can I draw this in? let's go over here.

What we end up with, I'm going to take this term, put it first. Negative B over 2A plus or minus taking the square term writing it first B squared minus 4AC square root of the 4A squared is just 2A. All right. Kind of ran out of room right here, but hopefully you can see that we have a common denominator.

So we can just throw that all together leaving us with negative B plus or minus the square root of B squared minus 4AC all over 2A.

A lot of work but I wanted to take some time and show you the quadratic formula is not some mysterious formula. It's just found by completing a square using the general quadratic of AX squared plus BX plus C.