 So how many mathematicians does it take to screw on a light bulb? The answer is 1. He turns the light bulb into three politicians thereby reducing it to the previous joke. And this is actually a joke that has some relevance here. Mathematicians like to solve problems by changing them into problems that they've already solved. So in general, one thing we can do is if we see an expression appear repeatedly in an equation, we might try as a substitution and see if we can reduce the equation to a whole number of power of that expression. So for example, let's take this equation square root of 3x-5 plus 12 equals 3x-5. And our observation powers say that, oh look, there's this 3x-5 here and also here. We see this expression appear repeatedly. So maybe we can do something with that. And if we can do some sort of a substitution, maybe we can reduce this thing into a more familiar type of equation. And in this particular case, what we might try is the substitution y equals square root of 3x-5. So this becomes y. This is y squared and this equation reduces to y plus 12, which is a quadratic equation and we already know how to solve quadratic equations. So if we don't know how to solve this, I can reduce it to a previously solved problem. Likewise, I might have a complicated six-degree equation here. And well, I don't know if I want to solve a six-degree equation, but the thing I might notice here is all the powers on x are even and every power of x is even. And what that suggests is that I might try the substitution y equals x squared. And that means y squared is x to the fourth, y cubed is equal to x to the sixth. And so what I have is a six-degree equation, but with this substitution I could rewrite it as a third-degree equation, which is going to be a little bit easier to solve. So let's see how we can apply that. So there's my equation involving the square root. I make the observation that 3x-5 appears a couple of times in this equation. If I let y equals the square root of 3x-5, well, again, the definition of square root. Square root is the positive thing whose square is the value there. So if I square both sides, y squared equals the absolute value of 3x-5. And dealing with absolute values often introduces complications. So, well, actually I check out. The clerk offers me the following deal. I can drop the absolute value. I can purchase the don't worry about the absolute value package as long as I verify my final solution. So the substitution allows me to transform the equation into something a little bit easier to work with. If y equals square root 3x-5, I have y plus 12 equals y squared. And now I have a quadratic equation, and so I can solve this. I'll get all the terms on one side, and I'll use the quadratic formula. y equals, well, a is 1, b is negative 1, c is negative 12, so I'll drop those into the quadratic formula. And after all the arithmetic does settles, I end up with my two solutions, 1 plus 7 over 2, or 1 minus 7 over 2, negative 3. So y equals 4, or y equals negative 3. And at this point, we apply the kindergarten rule. We put things back where we found them. So y equals 4, well, y equals square root 3x-5. So I have square root 3x-5 equals 4. And I do want to solve this equation eventually for x, so I need to get rid of that square root. I'll square both sides. And I have a nice simple equation that I can solve, and after all the arithmetic does settles, I've got my solution, x equals 7, there's one possible solution. The other one, y equals negative 3. Well, again, put things back where you found them. y was square root 3x-5, so I have square root 3x-5 equals 3. And I'm going to ignore that saying that we have a little bit of analysis goes a long way. I'm going to ignore any sort of analysis of this statement here. Now, why might I actually want to do that? Well, notice that I have square root equals a negative number. Specifically, this is a negative number, and I know that square root is the positive number and square is the inside, so I know that this should not give me a solution. But I'll ignore that fact, and I'll blindly rush ahead doing writing things down and doing the algebra. So I'll square both sides, I'll add 5, I'll get my solution, x equals 14 over 3, and there's a possible solution. Now, I did purchase at checkout the Don't Worry About Absolute Value Package, but the price we pay for this extended warranty is not really a very expensive thing to pay because we always, always, always, always, always should check our solutions. So we need to verify that they work in any case. This package is cheap at that price. So I'll check x equals 7. At x equals 7, again, I'm trying to solve this equation, not this equation, not anything else. I'm trying to solve this equation. If x equals 7, then I'll drop that into my original equation and see if I get a true statement. I'll let the arithmetic dust settle and let's see, squared at 16 plus 12, that's 4 plus 12, does in fact equal 16. So x equals 7 is in fact a solution. I also have to check x equals 14 thirds. Again, remember, this solution came from this original equation here and I had my doubts as to whether this equation would actually have solutions because it claims squared is equal to a negative number. I would not expect 14 thirds to actually give me a solution. But let's check it out. If x equals 14 thirds, dropping that into my original equation, I get this statement here and after all the arithmetic dust settles, I get this statement and 3 squared of 9 is 3 plus 12 equals 9 and this is a true statement, so x equals 14... Wait a minute, that's not true. 3 plus 12 is not equal to 9, so this is not a solution just as expected. So our solutions, our only real solution, is going to be x equals 7. How about this 6 degree equation here? So again, the observation that we can make here is that all of our powers of x are even and that suggests that x squared appears repeatedly because if y equals x squared, then my higher powers of y are also going to be even powers of x. So I can make this substitution, y equals x squared, y squared is x to the fourth, y cubed equals x to the sixth and my equation becomes, from a 6 degree equation, becomes a much simpler third degree equation. So now how do I find solutions to this? Well, the rational root theorem guarantees that the rational roots have to be among divisors of this divided by divisors of the coefficient. So that means if I have any rational roots at all, they have to be among plus or minus 1, plus or minus 2, plus or minus 3, 4, 6, or 12. So what do I do? I magically guess that the root must be 2. Well, actually, how do I have to do that? Well, I have to try out all those values and then eventually I find x equals 2, y equals 2 works as a root. There is no shortcut to finding out the root. We have to try all of our values until we find one or fail to find one. So y equals 2 does work out to be a root. My synthetic division algorithm tells me that it's a root and it also gives me the other factor. It tells me that y minus 2 is a factor and the other factor is y squared plus 5y plus 6. So I can factor the left-hand side times this equals 0. And so the zero factor theorem says that either y minus 2 is 0 or the other thing, y squared plus 5y is equal to 0. So again, I have my solutions, y minus 2 equals 0, y equals 2. Put things back where you found them. Again, that's our kindergarten rule. So y is x squared. So x squared equals 2. And so that tells me x is plus or minus square root of 2. The other sets of solutions come from the second factor equal to 0, y squared plus 5y plus 6 equals 0. And I can solve this. I'll use the quadratic formula and I find I have two solutions, y equals negative 2 or y equals negative 3. So again, put things back where you found them. So I have x squared equals negative 2. So that tells me x is plus or minus i square root of 2. y equals negative 3. Put things back where you found them. x squared equals negative 3. x equals plus or minus i square root of 3. And here are our 1, 2, 3, 4, 5, 6 solutions to this 6-degree equation. Now one quick observation. If you had plunged ahead and tried to find the rational roots of this, you wouldn't have found them because there aren't any. Even though this is a polynomial that we could potentially apply the rational root theorem to factor, the problem is that none of the actual roots are rational. The only way that we can actually solve this is we have to do this substitution that's going to reduce this to a cubic equation. So the important lesson here is do a little bit of analysis before you start to solve a problem. To make the observation that we can reduce this to a third-degree equation, we should reduce this to a third-degree equation because it will allow us to get our solutions.