In some cases, an equation will involve several radicals, and not just one. Let’s take the case of two radicals. There’ll be two types of equations in this category. The first one is that when you see the m-th root of f equals the n-th root of g. See how carefully that has to be pronounced? If the functions, f and g here, are defined on a domain capital omega, then of course, the natural domain for our equation will restrict the x to omega. But also it will include the conditions that f and g be positive, because otherwise the roots are not defined.
What we’d like to do is raise each side of the equation to a certain number so that the radicals disappear. And here’s what we’re going to have an equivalence, for x and D of our original equation, with the equation f to the n equals g to the m. Can you see what power we raised the terms to, in order to arrive at this? Answer, we raised them to the power m times n. Let’s look at a simple example. We want to solve square root of x equals cube root of 2 minus x. The natural domain for this function, well, we want x to be positive. We want 2 minus x to be positive.
So the interval 0, 2 is the natural domain. Now, if x is in that interval, the equation is equivalent to this one. To what power have we raised the original equation terms in order to arrive at this? Yes, we’ve raised them to the power 6. Now, we can work on this equation, this new equation that we’ve generated, and we arrive at a cubic equation for x. How are we going to find the roots of this cubic polynomial? Answer, we’re going to use ingenuity, ingenuity and factoring.
We’re going to take an x squared common factor out of the first two terms and a 4 out of the last two terms, and now there arises a common factor to both of these terms, the x minus 1. When we factor it out, we’re left with x squared plus 4, which is an obviously irreducible quadratic, and therefore, the only route is x equals 1. We go back. We check that 1 is in the domain. We check that it satisfies the equation, and the solution set, therefore, is the singleton 1. Now, we’re going to look at the presence of more than one radical, but in a different way.
You see that on the left hand side, we have the sum of two radicals, and on the right, we have one, or we could not have one, it would be much the same. So the functions f, g, and h defined on omega. The natural domain would be one which includes the restrictions that these functions are positive, for evident reasons. Given that we’re in the domain, we can now square both sides of the equation, and when we work out the square, on the left hand side, by a notable identity, you will see that we still have a radical. But now, there’s only one radical left in the equation.
So it’s the type of equation we dealt with earlier, which we can now go and proceed to solve the way we did. Now, this new equation, with a single radical, is not one that you’re supposed to memorize a formula for, of course. It’s a matter of retaining the method, not actually any kind of formula. Let’s apply this reasoning to solving this typical equation involving two radicals. What do we do first? Well, we put one of the radicals alone on one side of the equation for simplifying the arithmetic later. We square the equation. Now, do we have an equivalent equation? Well, let me confess to you, I’m going to do it here in cowboy mode.
I’m not even going to think of the domains, I’m just going to go full speed ahead. We still have one radical, so we isolate it on one side of the equation. We square, again a second time, the equation, and now all the radicals are gone. We simplify the equation. We wind up with a simple quadratic one whose roots are evident, 0 and 8. We check whether or not they are solutions. We need to check. Checking is always a good thing, but remember when we square equations, we might introduce extraneous solutions that aren’t really solutions. So checking goes from being a good idea to being an absolutely obligatory idea. Is 0 a solution? We plug it in. Does that work? Nope.
We try x equals 8. Does that work? Well, we do the arithmetic. We find, yes, it works. So the solution set is the singleton 8. Let’s look how change of variables, which we’ve seen before, can also work in certain radical equations. Here’s one where you see there are two radicals, x to the 4/3 and x to the 2/3. The domain, natural domain of this equation, is R+, 0 infinity. Now, we’re going to set t equal x to the 2/3. That’s going to be our change of variables. How did we think of this change of variables? Well, we looked at the equation, and we desperately searched for something that would make it simple, and this is what we came up with.
What is the equation, the original equation in terms of this new variable? Well, it’s a very nice one, t squared minus 5t plus 6 equals 0. It factors t equals 2 and t equals 3 are the roots. Now, we have to go back, though, to the original variable. That was x. Remember that t is x to the 2/3. So when t is 2, that gives us x to the 2/3 equals 2, and when t is 3, we get x to the 2/3 equals 3. And that generates two possible values for x, and this gives us our solution set. Next, we’re going to treat equations that have absolute value in them.