 So now that we know how to solve quadratic equations, what else can we do with them? Well, a lot of problems in mathematics can be solved by referring to what we might call the kindergarten rule, a basic rule that you learned back in kindergarten. You can take something out as long as you put it back. So for example, let's consider an equation like this, 7x plus 5 squared plus 8 times 7x plus 5 plus 15 equals 0. And so the thing we might notice here is that we have a 7x plus 5 appearing in two locations. So suppose I take it out. Now, here's an important idea to keep in mind. This 7x plus 5 and this 7x plus 5 are the same thing. They both are 7x plus 5. And so that means if I take the 1 out, I really need to take both of them out because they are the same thing. And this is sort of a variation on a slightly different kindergarten rule. You don't get to treat identical things different ways. So I take out the 7x plus 5, they're gone. And the thing I might notice here is this now looks like a quadratic equation. This is something squared plus 8 times something plus 15. And my quadratic formula works very nicely to solve equations like that. So I have A equals 1, B equals 8, C equals 15. And I can drop those into my quadratic formula. And I can follow the arithmetic. I'll let the arithmetic does settle. And at the end of it, I'll have my two solutions, negative 8 plus 2 over 2, negative 6 over 2, negative 3, or negative 8 minus 2 over 2 minus 5. And so I have two solutions to whatever this was. And now here's where the kindergarten rule applies. We have to put everything back where we found it. So we took out the 7x plus 5, we have to put it back in every place that it should have been. We'll drop it back into there. And what is important here is we actually now have two equations, 7x plus 5 equals negative 3 or 7x plus 5 equals negative 5. And now I can solve each equation individually. 7x plus 5 equals negative 3. I'll solve that equation. And I'll get x equals negative 8 7s as one solution. I have my second equation, 7x plus 5 equals negative 5. I'll solve that one. And I get x equals negative 10 over 7 as my second solution. And this allows us to generalize this problem of solving the quadratic. In general, I can say that an equation is reducible to quadratic if I can make some sort of substitution that transforms it into a quadratic equation. So as we saw, this equation 7x plus 5 squared plus 8 times 7x plus 5 plus 15 is reducible to quadratic because what I can do is I can use the substitution y equals this. I can take out 7x plus 5 and replace it with y. And I get the equation y squared plus 8y plus 15. I get a nice quadratic equation. Or I can take another equation, z to the fourth minus 4 z squared. That's also reducible to quadratic because I can take out a z squared, replace it with a y. And I get y squared minus 4y equals 12. And there, again, is a nice quadratic equation. And I can take the equation x cubed minus 3x squared plus 5x equals 7. And well, actually, I can't reduce that to quadratic. And there's a couple of different reasons why that's not going to be possible. The simplest, the most obvious reason here is that if I have a real quadratic equation, I can only have three distinct types of terms. I have a square term, my linear term, my constant term. Here I have four different types of terms, a square, linear, constant, and a cubic term. So I'm not going to be able to reduce this one to a quadratic. Now, if I want to solve an equation that's reducible to quadratic, the first thing I'm going to do is I'm going to, well, reduce it to a quadratic. I'll make some sort of substitution that gives me a quadratic equation. I know how to solve quadratic equations, so I'll do that. And then we'll apply the kindergarten rule, we'll put things back where we found them, and then we solve for the original variable. So for example, let's take z to the fourth minus 4z squared equals 12. And so the observation that we might make here is I have a z squared and a z to the fourth. I have even powers of z. And what to suggest, because z to the fourth is z squared squared, that if I make the substitution for y equals z squared, I might be able to reduce this to a quadratic equation. So I'll make the substitution y equals z squared. And let's see what happens. So here's my equation. Again, z to the fourth is z squared squared minus 4z squared. The equal says any time I see a z squared, I can replace it with a y. So I can replace the z squared with a y. I can replace the z squared with a y. And I'll do that substitution. And behold, I have a quadratic equation. And I know how to solve those. So I'll get all the terms onto one side, get the equation y squared minus 4y minus 12 equals 0. I'll take a note of what I did. I'll let y equals z squared, my equation becomes this. And now I can use the quadratic formula to find the two solutions. So I have the expression in the correct form. A equals 1, B equals negative 4, C equals negative 12. I'll drop that into the quadratic formula. And after all the arithmetic does settles, I end up with my two solutions, 6 or negative 2. So letting y equals x squared into this equation, I have my solutions y equals 6, y equals negative 2. And now we apply the kindergarten rule. y is 6, y is z squared. Put things back where you found them. z squared equals 6. And I can solve that. And the other one, y equals negative 2. Put things back where you found them. y and z squared are the same things. z squared equals negative 2. And so z is plus or minus square root of 2. I root 2. And there's my solution.