# Video: Solving Quadratic Equations by Factorization

Find the solution set of 𝑥⁵ − 33𝑥^(5/2) + 32 = 0.

03:36

### Video Transcript

Find the solution set of 𝑥 to the power of five minus 33𝑥 to the power of five over two plus 32 equals zero.

Well, with this type of question, what we can do is first of all take a look at the powers we’ve got of 𝑥. So, if we look at the exponents, we can see that we’ve got five and five over two or five-halves. Well, if we look at 𝑥 to the power of five over two and then we square it, what we’re gonna get is 𝑥 to the power of five. And that’s because with our exponent rule, what we do is we multiply them. So, five over two multiplied by two gives us five.

So therefore, what I’m gonna do is rewrite our equation. And I’m gonna say, first of all, that 𝑎 is equal to 𝑥 to the power of five over two. So, we’re gonna get 𝑎 squared — and we get 𝑎 squared because we’ve already explained why; and that’s because if you square 𝑥 to the power of five over two, you get 𝑥 to the power of five — minus 33𝑎 plus 32 equals zero.

Now, if we factor this to enable us to solve it, what we’re gonna get is 𝑎 minus 32 multiplied by 𝑎 minus one equals zero. And we get that because we’re looking for two factors whose product is positive 32 and whose sum is negative 33. Well, if we’ve got a negative sum and a positive product, then we know both of the factors need to be negative. And negative 32 multiplied by negative one is positive 32, and negative 32 add negative one gives us negative 33.

So, we know these are our factors. So therefore, our solution is gonna be 32 or one. And we know this is the case because one of our parentheses needs to be equal to zero. And if we had 𝑎 minus 32 and made it equal to zero, then if we add 32 to each side, we get 𝑎 is equal to 32. And if we had 𝑎 minus one equals zero, then if we add one to each side, we get 𝑎 equals one which is what we said. So, great, we found our solution.

So, does this mean that we’ve solved the problem and solved our quadratic equation? Well, no because we’ve only found 𝑎 and what we wanted to find is 𝑥. So, what we do now is we look back at our original substitution, which was 𝑎 is equal to 𝑥 to the power of five over two. And now, what that means is what we’re gonna have is 𝑥 to the power of five over two is equal to 32 or 𝑥 to the power of five over two is equal to one. Now, to solve the left-hand equation, we’re gonna use a couple of exponent laws.

The first one is that root 𝑥 equals 𝑥 to the power of a half, and the other one is that 𝑥 to the power of 𝑎 over 𝑏 is equal to the 𝑏th root of 𝑥, and this is all to the power of 𝑎. So, by applying both of our rules, what we’re gonna get is root 𝑥 to the power of five equals 32. So then, if we take the fifth root of both sides of the equation, what we’ll get is root 𝑥 is equal to two. And then, if we square both sides of our equation, what we’re gonna get is 𝑥 is equal to four. So, that’s one of our solutions.

Well, if we move on to the right-hand side, this is a lot easier to solve because we’ve got 𝑥 to the power of five over two equals one. Well, we know that one to the power of anything is one. So therefore, we know that the other solution is 𝑥 equals one. So, we could think, well we finished. But actually, if we look back at the question, we’re looking for a solution set. So, we want to use set notation. So therefore, we can say that the solution set of 𝑥 to the power of five minus 33𝑥 to the power of five over two plus 32 equals zero is one, four.