### Video Transcript

Find the solution set of π₯ to the fourth power minus 25π₯ squared plus 144 equals zero in the set of real numbers.

Now at first glance, this may look like quite a complicated equation. The key here though is to spot that we can rewrite π₯ to the fourth power as π₯ squared all squared. And this is because when we have an expression of this form, we simply multiply the exponents. So π₯ squared squared is π₯ to the power of two times two, which is π₯ to the fourth power.

And then if we look really carefully, we might spot this looks a little bit like a quadratic equation. And so to solve this equation, we perform a substitution. Weβre going to let π¦ be equal to π₯ squared. Then π₯ to the fourth power, which is π₯ squared all squared, becomes π¦ squared. And then our earlier equation becomes π¦ squared minus 25π¦ plus 144 equals zero. Now we see we do indeed have a quadratic equation. So how do we solve this?

Well, one way is to factor the expression on the left-hand side. We have the product of two binomials. And the first term in each must be π¦ since π¦ times π¦ gives us π¦ squared. And then we need to look for two numbers whose product is 144 and whose sum is negative 25. Well, the only two values for which that is true are negative 16 and negative nine. So our equation becomes π¦ minus 16 times π¦ minus nine equals zero.

Now π¦ minus 16 and π¦ minus nine are themselves simply numbers. And their product is equal to zero. So what can we say about each of those numbers? Well, for that to be true, for their product to be equal to zero, at least one of those numbers must themselves be equal to zero. So either π¦ minus 16 is equal to zero or π¦ minus nine is equal to zero. And we can solve each of these equations for π¦.

With our first equation, we simply add 16 to both sides. And we find π¦ is equal to 16. And similarly, with our second equation, weβll add nine. So we get π¦ is equal to nine.

Weβre not finished though. Remember, we were solving an equation in π₯. So we need to replace π¦ with π₯ squared as in our earlier substitution. And in doing so, we create two further equations. We have π₯ squared equals 16 and π₯ squared equals nine. Weβll solve each of these equations for π₯ by square-rooting both sides. Remembering that we need to take both the positive and negative square root of 16 and nine.

Well, the square root of 16 is four and the square root of nine is three. So π₯ is either positive or negative four or positive or negative three. And we can use these little curly brackets to represent the solution set. The solution set of π₯ to the fourth power minus 25π₯ squared plus 144 equals zero is four, negative four, three, negative three.

And of course, we could check these solutions by individually substituting into the expression π₯ to the fourth power minus 25π₯ squared plus 144. If we did that, we would indeed get a value of zero as required.