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The case of two radicals

The case of two radicals
Suppose now that two radicals are present in an inequality. How do we proceed to solve it? Well, you know, the overall strategy is similar to the case of equations. Specifically we’re going to separate the radicals, put one on one side, one on the other. We’re going to raise to a certain power to eliminate one radical. There will still be one left usually. We will isolate that one remaining radical on one side and raise again to the appropriate power, and we’ll obtain after this a de-radicalized, so to speak, inequality, no radicals. And we’ll solve that one, we hope.
And the difference, though, as we’ve pointed out, is that you really must be careful to record along the way the domain and the various equivalence conditions that we obtain. Otherwise you won’t be able to figure out the correct answer. Let’s illustrate this in the case of this inequality, which we wish to solve. We begin by noting the natural domain. We want the square roots to be defined. So x plus 1 has to be positive and 3x plus 1. That amounts to saying that x has to lie in the interval minus 1/3 infinity, the natural domain. Now we separate the radicals.
One has gone over on the left and the other on the right, and you notice that I’ve kept the plus 2 on the left side. Why have I done that? Well, because if I do that, the left side is clearly a positive number, as is the right side. And so when we take the square of both sides, there will be no further restrictions on x to record. If the 2 were over on the other side as a minus 2, there would be further restrictions. We could still note them and proceed and get the right answer, but this seems a little simpler. Now, in the domain we can square both sides of our latest inequality and get something equivalent.
And when the dust settles, we’ll put the one remaining radical over on the left. And now we wish to square both sides again to get rid of the remaining square root. Now, however, there will be an equivalence condition when we do that. The condition that x has to be greater than 2 in order for the right side to be positive. So we note this carefully for later use. So squaring both sides in the previous inequality and letting the dust settle, we finally arrive at a completely de-radicalized inequality. It’s a simple quadratic inequality. And we see that the solutions are the x’s that lie left of 0 and right of 8.
Now, we can’t stop there because we can only keep the x’s that we’ve just found that are compatible; that is, that lie in the domain and satisfy the equivalence conditions, whatever they were, that we got along the way. In this case, they have to be greater than 2. And we see, therefore, that the ones we keep are those in the interval 8 infinity, which is our solution set. Let me remark that if you are able to solve this kind of inequality, then you are among the elite of inequality solvers. And I congratulate you.
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Precalculus: the Mathematics of Numbers, Functions and Equations

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