 Given a quadratic equation, we can solve for x using the quadratic formula. We often encounter equations that we can transform into a quadratic by a change of variables. A useful idea equals means replaceable. If we introduce a new variable, we might be able to rewrite a more complicated equation as a quadratic equation. For example, consider this equation. While we could expand and collect like terms, it would be a lot of extra work. Instead, notice that our variable k appears in the expression 1 plus k and 1 plus k squared. So if x equals 1 plus k, then x squared is 1 plus k squared. And using our substitution, x equals 1 plus k gives us... ...and rewriting our equation in standard form. And now we can apply the quadratic formula to get our solutions. Now, it's important to remember that we've actually solved for x, and so remember, put things back where you found them. We actually went to solve for k. So x was equal to 1 plus k, so we'll replace and solve. The key here is recognizing when we can do a substitution that will give us a variable and the square of that variable. So in this equation, we might notice that v to the fourth is the same as the square of v squared. And so we can substitute x for v squared and x squared for v to the fourth. And using the substitution gives us... ...and again we have a quadratic equation which we can solve. Again, we'll restore our original variables. Now, we can solve v squared equals 8 by extracting the square root of both sides. However, v squared equals negative 1 has no real solutions. And so this doesn't give us a new set of solutions, and our only two solutions are plus or minus the square root of 8. Or let's consider a more complicated equation. So if we let u equal 1 divided by 1 plus x squared, then u squared is 1 divided by 1 plus x to the fourth. And also 5u is this term 5 divided by 1 plus x squared. And so using the substitution gives us a quadratic equation we can solve. And it's always helpful to break apart the plus or minus into the two actual solutions it represents. Again, putting everything back where we found it, we know that 1 divided by 1 plus x squared is 5 plus square root 13 over 2, or 5 minus square root 13 over 2. Now, we still want to solve for x. So let's take one of these equations. Now we can verify in a calculator that the right hand side is positive, so we can take the square root of both sides. And this gives us where because we're taking a square root, we need to make sure that we include both the positive and the negative square roots. Taking the reciprocals and solving gives us two solutions. And we'll get another pair of solutions from our other equation. Thanks for watching.