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Inequalities with one radical

Inequalities with one radical
We’re now going to discuss inequalities featuring a radical. We’re going to start with some general remarks on strategy, which turn out to be rather important. Because we can miss out on solutions, we can get a wrong answer, if we work insensitively. I’ll explain what I mean. Just like for equations, although we love radicals, we try and get rid of them. That is, we, at some point, try and raise the inequality to the same power on both sides in order to eliminate the radicals to be able, then, to find a solution set.
The only really viable approach here is to surgically dissect the domain as you go, to take account of the domain and the equivalence as you proceed to perform these operations, and at the end arrive at the correct solution set. We’ve seen an example early on in this week that neglecting the domain and operating in the cowboy procedure definitely won’t work in general, it could give you a false answer. So let’s look at inequalities, with the first general type being the following. N-th root of f, less than g. The inequality here, again, can be strict, or it can be non-strict. It doesn’t make a big difference in how you solve them.
The inequality will have a domain and, naturally, the domain will be contained in the set of x’s for which f is positive. Why? Because the n-th root of f is only defined if f of x is greater or equal 0. Now, we’re going to study what happens when you take the n-th power of each side of the inequality, and whether we get equivalence and in what sense. For x in the domain, here’s the equivalence. Our original inequality is equivalent to the one in which we’ve taken the n-th power of each side, so we get f less than g to the n. However, you have to also take account of the restriction that g of x is positive. Why?
Because if the n-th root is going to be less than g, then the n-th root being positive means g has to be positive. Now, this equivalence is easy to prove. You could try it as an exercise. Just show that if x is in the left hand side– that is, satisfies the inequality on the left– then it must satisfy the two conditions on the right, and vice versa. If x satisfies the two conditions on the right, it satisfies the inequality on the left. So we have an equivalent inequality. Now, in the second case, you have an inequality of the form n-th root of f, greater than g, rather than less than.
You might think it’s rather similar, but in fact, it’s pretty different, because the equivalence now is the following. When you take the n-th power of each side, you get a new inequality, f greater than g to the n, and anything solving that will solve the original. But also, anything for which g of x is negative, any x for which g of x is negative, also solves the original inequality. That’s fairly evident. So, in the first case, the g of x being positive was a restriction that will make your set of solutions smaller. You have to cut away those points that fail to satisfy that.
In the second case, the word “or” indicates that it will give you more solutions that you mustn’t lose track of. Now, all of this will be much clearer when we do an example of each type. Let’s do that now, starting with the first type, n-th root of f less than g. So you can see this is of that form. It’s the square root of x plus 3, which must be less than or equal x plus 1. We’re sensitive to the domains in our approach, so we immediately identify the natural domain, which is the interval minus 3 infinity for evident reasons of the square root being defined.
Now, what about the equivalence when we square both sides to get rid of the radicals sign? Well, we will have equivalence, so the new inequality will be x plus 3, less than x plus 1 squared, except that we mustn’t lose track of the extra restriction that x plus 1 has to be positive. So we don’t lose sight of that fact, although we temporarily forget it in solving the squared inequality. The squared inequality, when the dust settles, is a quadratic inequality whose solution set is easily seen to be the union of two intervals, the points up to minus 2 and the x’s beyond 1. So there is the solution set of our squared inequality.
But now we have to remember our restriction. We have to see whether the x’s we’ve found, or which x’s we’ve found, are compatible. That is, compatible with being in the domain, D, and also compatible with the restriction we identified above, that x has to be greater than minus 1. When we take only those points, we find the interval 1, infinity. And that is, in fact, the solution set of our inequality. Now, should we check this solution set? Well, if we’re not androids, yes, we should. But notice that the word “check” here has a really different meaning from when we were doing equations.
For equations, you had two or three values, you go back and see if they work, you reject the ones that don’t work. But here, it’s really rather different. You can’t do that, and checking means checking your work carefully over again, which is always a good idea, of course. Next on the agenda is an inequality of the second general type that we were discussing. You know, the n-th root of f greater than g, rather than less. So this example is the square root of x minus 2, strictly greater than x minus 4. Natural domain, for reasons of the square root being defined, the interval 2, infinity.
When we square the inequality on both sides to get rid of the radical, what will be the equivalence statement? It’ll be this. That the new points we’re looking for are those which satisfy the squared inequality, or which satisfy x minus 4 being negative. We mustn’t forget those points, otherwise they will be missing in our final solution. So we look at the squared inequality, we analyze it, it turns out to be a quadratic inequality. Easy to analyze, and its solutions consist of the open interval 3, 6. Now we have to talk about compatibility and also the “or” part above. We want to keep the solutions that we’ve just found, the interval 3, 6, but only those that are in the domain.
Well, they happen to all be in the domain, so we keep them. But now we have to add the “or” part. That is, we have to take account of the points x less than 4, but only those that are in the domain. So that gives us the interval 2, 4. And now the union of these two intervals that we’ve identified will be our overall solution set. The union turns out to be, as you can see, the interval 2, 6, half open. Let me remind you of the complementarity principle. We’ve just solved this inequality relative to the natural domain, and we’ve found the interval of solutions to be 2, 6.
Suppose someone now asks us to solve the opposite inequality, where the greater than is replaced by less than or equal. We don’t have to redo the work, we know the answer. It’s the complement of the original solution set relative to the natural domain, which turns out to be the closed interval 6, infinity. Our next topic will be to consider inequalities in which there are two radicals.
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Precalculus: the Mathematics of Numbers, Functions and Equations

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