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2.15

## Precalculus

Skip to 0 minutes and 11 seconds Simplifying radicals is an important process in mathematics, and it requires some practise to do even if you know all the laws of radicals and exponents quite well. Let’s look at some examples of how this can arise. Here’s the function defined by the defining formula you see. We wish to simplify this function, and at the same time, determine the natural domain of the function. As regards to the domain, we realise that since radicals are only defined for positive numbers, we have to have x plus 3 cubed positive. That amounts to saying that x plus 3 has to be positive, and therefore, the natural domain of our function is the interval minus 3 to infinity. Now what about simplifying it?

Skip to 1 minute and 6 seconds Well, for any x in the domain that is greater than minus 3, we can write this formula. And now we can cancel out the 3. How have I done that? Well, you get down to this by using this formula from the past. And you see that there should be absolute values around the x plus 3 strictly speaking. But x plus 3 is positive since x is in the domain. And therefore, we don’t need to put the absolute values. And therefore, we have the answer to our problem. We’ve simplified the function. We have defined its domain. Another very common type of exercise in working with radicals is the so-called procedure called rationalising the denominator.

Skip to 1 minute and 56 seconds This means removing radical signs from the denominator to get an equivalent but different expression for the number. Now, if your fraction is of the type a over the n-th root of b, then it turns out to be a very useful trick to multiply both the top and the bottom of your number by the n-th root of the n minus first power of b. You’ll see why this trick is so useful. It does the job. Now a trick means that somebody had a good idea once. And we remember the good idea even if we don’t remember the somebody.

Skip to 2 minutes and 36 seconds The reason this trick is going to work is because in the denominator, when we get the product of these two terms, it’ll just reduce to b. Let’s look at an example. It will clarify the whole procedure I think. Here’s 1 over the cube root of 7. We wish to rationalise the denominator, get rid of the radical sign in the denominator. Well, we can multiply across by the cube root of 7 squared over itself. That changes nothing because we’re multiplying, essentially, by 1. However, when we do that, the denominator becomes the cube root of 7 cubed, which is 7. And the numerator stays the way it is, and we have rationalised the denominator.

Skip to 3 minutes and 21 seconds Here’s another instance of the technique called rationalising the denominator. But now, in the case when the denominator is the sum or the difference of two radicals. The very useful trick, which should be remembered, in this instance is to multiply across by an expression like this. The reason– notice the sign by the way is reversed in this. If we had plus in the original thing, then it’s negative in the trick device, and vice versa. The reason this will be useful is because in the denominator, when we multiply, for example, root a minus root b by its conjugate, as this is called root a plus root b, well, when you work that out, remove the parentheses. You can check.

Skip to 4 minutes and 7 seconds You get a minus b. No radicals. Using a simple example to illustrate the procedure, rationalise the denominator in this number. The trick involves multiplying across by what you see. In the denominator, this winds up giving 5 minus 3. That is, two. The numerator is what it is, and we get the final answer in which there are no radicals in the denominator. The last, and perhaps, more difficult example, I ask the question is it true that these two numbers that you see are the same? In calculus, you’ll sometimes calculate a length or an area, and you find by doing it two different ways, you get two different answers. This is actually an instance of that.

Skip to 4 minutes and 58 seconds And you want to know if your answers are the same, which they should be. How do you prove they’re the same? Well, first, you could check that what’s inside the square root sign is positive, which it should be. In fact, that amounts to checking that 13 is greater than 4 root 3. That’s equivalent to checking that 13 squared is greater than the square of 4 root 3 because we’re dealing with positive numbers. And the function x squared is a strictly increasing function on r plus. When you work that out, it amounts to checking that 169 is greater than 48, which it is. So we’re taking the square root of a positive number. That’s good.

Skip to 5 minutes and 38 seconds Next, you should check that the right hand side is positive. Well, that amounts to checking, ultimately, as you can see, that 12 is greater than 1, which it certainly is. So we have two positive numbers, which may or may not be the same. However, that amounts to saying that when we square both sides, they will be the same. Now, when we square both sides, on the left, we get exactly 13 minus 4 root 3 because we’ve checked that that’s positive. When we square on the right hand side, well, you get this parenthesis times itself, you work it out with the distributive law and so forth, and you’ll get an answer, which is the same as on the left hand side.

Skip to 6 minutes and 21 seconds So we have proved that those numbers are the same.