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4.12

## Precalculus

Skip to 0 minutes and 11 seconds Let’s look for radical solutions, or rather solutions to equations that contain a radical. These arise rather often in applications. Here’s our general equation, the nth root of f equals g. Now, we’re tempted to identify the domain, because we’ll see that it has some interest, and the natural domain here would be one that includes a restriction on f being positive. Why? Because the nth root is only defined for a positive number. We’re also tempted to take the nth power on each side of our original equation, because that will get rid of the radical and possibly make it easier to deal with.

Skip to 0 minutes and 56 seconds Now, bear in mind that when you have positive numbers, then a and b, a and b are equal, if and only if their nth roots are equal. So that’s fine, but you must remember that this is for positive numbers. So when you take the original equation, and you raise each side to the nth power, you get f equals g to the n, but in doing so, you have to remember to take account of the condition, g of x greater than or equal 0, in order to have fully equivalence of the left hand side and the right hand side.

Skip to 1 minute and 28 seconds Now remember, equivalence means if x satisfies the equation on the left, then it satisfies both conditions on the right, and vice versa, if x satisfies both conditions on the right, then it satisfies the left hand side. A common error is to assert that nth root of f equals g is the same as f equals g to the n. This is simply not true. When n is even, the new equation that you have generated will usually have more solutions, solutions for which g of x is negative. Let’s look at an example to clarify this. We want to solve the equation root x plus 3 equals x plus 1. The domain of this equation?

Skip to 2 minutes and 13 seconds Well, we might say that it’s the interval minus 3 to infinity, because we want x plus 3 to be positive, since it’s under the square root sign. Now, equivalence would mean that we could square both sides, and we do this here provided we bear in mind the condition, as I just said, that x plus 1 will have to be positive. You’ll notice that this introduces a further restriction on x. If we were very insightful, we might have included that further restriction right away in the initial definition of the domain. But it doesn’t matter, we note it, and we pursue the study.

Skip to 2 minutes and 53 seconds We look at the new equation, we work on its terms a little bit, and we see that it’s a quadratic equation, which we can easily solve, the roots are 1 and minus 2. Now can we stop at this point? No, we really need to check the compatibility of these candidates with the domain conditions that we have. Now, these compatibility conditions are going to lead us to reject x equals minus 2. Why? Because minus 2 does not satisfy x greater or equal to minus 1. On the other hand, the other candidate, the 1, that’s fine, it satisfies all the compatibility conditions. We check it out in the original equation. Naturally, we see that it works, and we have our answer.

Skip to 3 minutes and 42 seconds The solution set is the singleton set 1. Now step back and look at how we solved this problem. We applied a sort of surgical approach. We dissected the equation, we cut away the irrelevant parts of the domain, and we were sensitive to the domain throughout. It turns out that there’s another way to attack such problems. It’s the insensitive approach, you might say, I call it the cowboy approach. We’re going to ignore the domains as we work. We’re going to go fast. We’re not going to check anything. We’re going to get some candidates, and then we’ll simply come back and see if they work or not. Let’s solve an example in this cowboy mode. Here’s an equation we wish to solve.

Skip to 4 minutes and 27 seconds How do we begin? Well, the first step here would be to isolate the radical on one side of the equality by itself. This will simplify the resulting arithmetic. Now we square both sides. There we go. We don’t even think about the domains or equivalents. We massaged the equation to make it look simpler. We come across a quadratic equation. It factors, therefore, the roots are minus 3 and 6. And so we have this conclusion, which is what? Well, it’s a necessary condition only. What it is, is the following conclusion. If x is a solution of the original equation, then it is necessary that x be 6 or minus 3, but it doesn’t work the other way. It’s not sufficient.

Skip to 5 minutes and 12 seconds We need to go back and see whether they are solutions. It’s quite possible that one or both are not, so we check. We plug in x equals 6 into the original equation. Does it work? Nope, doesn’t work. Well, we reject 6. We put x equals minus 3 into the equation. Does that work? Well, yes it works, therefore, we keep x equals minus 3. Our solution set is the singleton set minus 3. Now that was the cowboy approach. You see how fast it was? It can be faster, and it involves less thinking. But it is true that the domain sensitive approach, the surgical approach, can also save time.

Skip to 5 minutes and 58 seconds In fact, in mathematics, it’s often the case that a little thinking at the beginning can save quite a bit of work later on. Consider this example. If we worry about the domain, we see that x should be greater than 1/2 for the square root to be defined. On the other hand, the square root of something is positive, so minus x is going to be positive, and these two conditions, we immediately see, are contradictory, so there are no solutions of this equation. So it would be unnecessary to go ahead and square and come up with a false possibility that we would then eliminate. Now, the question arises, is it better to be a surgeon or a cowboy?

Skip to 6 minutes and 38 seconds We need not resolve this question immediately, but it will arise again pretty soon. Here’s another example of solving equations. Comes up a lot in physics, engineering, chemistry, economics, and so forth. You have a formula in which parameters appear, and you wish to interchange parameters in the formula. Let’s take an example. In mechanics, when you’re looking at the oscillations of a pendulum in the plane, there is a formula that expresses the period of the oscillation– that’s the time it takes to make one full swing back and forth– the period capital T, as a function of the length of the pendulum.

Skip to 7 minutes and 21 seconds And the formula then is T equals 2 pi square root of L over g, where g is a certain positive constant. Now, in this formula, we’re thinking of L as an independent variable, which, once you have its value, determines the value of T, the period through the formula. However, it’s possible to think of things in the opposite way. We could imagine that you choose the period first, and then you have to find the length that will give you that period. In other words, we’d like to express L as a dependent variable of the independent variable T. You just solve the equation for L. Here’s what we do.

Skip to 8 minutes and 2 seconds First of all, I’ve switched the sides around, because I’m going to be writing at the end L equals, and I like my dependent variable to be on the left hand side of the equation. Now, all the quantities are known to be strictly positive, L, T, g, pi, and so forth. So we don’t have to worry too much about equivalence, and whether we’re getting extraneous solutions. We just go ahead and square both sides. And now we multiply by g and divide by 4 pi squared, and there is a formula for L. So, you know, when engineers do this kind of thing, they’re not being Cowboys. Well, not usually.