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Equations involving several radicals in practice

Equations involving several radicals in practice
Hello, welcome back to a step in practice. We are involved here with equations with several radicals. For instance, in exercise one, we are dealing with the equation, square root of 4 minus x plus the square root of x minus 2 equals to 2. First of all, let us study the natural domain of the equation. We must impose that 4 minus x is greater or equal than 0. x minus 2, greater or equal than 0. So x must be less than 4 and greater than 2. So the domain is the interval, the closed interval, 2, 4. So let us remind that we must have x between 2 and 4. Let us write it apart.
Now when x is in the domain, we can write the equation as square root of 4 minus x equals 2 minus square root of x minus 2. And then we proceed by implying some conditions. And we try to eliminate at least one of the roots. So this implies, for instance - but it is not equivalent, be careful. We may obtain more solutions than what we need - that the square is equal, the square of both terms is equal. So that 4 minus x equals 2 minus square root of x minus 2 to the square, in general it is not equivalent.
So what we shall do at the end is to check that our solutions are actually solutions to the initial equation. So this gives 4 minus x equals 4 plus x minus 2 minus 4 square root of x minus 2. So which is 4 times the square root of x minus 2 equals to 2x. And then we have 4 minus 2, which leaves 2, minus 2. You can simplify by 2 and we get 2 times the square root of x minus 2 equals x minus 1.
And then again, this implies for x greater than 2, and this is the case, that by taking the square of each term, that 4 times x minus 2 equals x minus 1 to the square, which gives x to the square minus 2x plus 1 equals 4x minus 8, and which is x squared minus 6x plus 9 equal to 0. Let us find the discriminant. The discriminant of the equation is 36 minus 36. So it is 0. And actually, we realized that this is a square. This is exactly x minus 3 to the square equal to 0. Actually, the only root of the polynomial is 3. And this gives x equals to 3. Now, compatibility, yes?
3 belongs to the interval 2, 4. But we also have to check that 3 is actually a solution because we proceeded by implication, not by equivalence. So we check that 3 is a solution, well the square root of 4 minus 3 plus square root of 3 minus 2 gives square root of 1 plus square root of 1, so 2. And on the other side, well, we get 2. Yes. The equation is satisfied. So the solution is 3.
In exercise 2, we consider the equation square root of 2x minus 8 equals square root of 3x minus 9. First of all, the natural domain of the equation is the set where square root of 2x minus 8 is defined, that is 2x minus 8 is greater or equal than 0 and where 3x minus 9 is positive.
That is the set of x, where x is greater than 4 and x is greater than 3. So the interval from 4 to plus infinity. For x in D, well, these two terms are positive. So the equation is equivalent to the square of the equation, 2x minus 8 equals 3x minus 9. And this is equivalent to the initial equation, which means x equals 1.
However, now we look at the compatibility condition, here we need that x is greater or equal than 4 and 1 is not greater or equal than 4. So we have to exclude 1. And you see, if you take 1, you get here minus 6, square root of minus 6 is not defined, actually. So the solution is the empty set. And this ends this step in practice. See you at the next step.

The following exercises are solved in this step.

We invite you to try to solve them before watching the video.

In any case, you will find below a PDF file with the solutions.

Exercise 1.

Solve the equation (sqrt{4-x}+sqrt{x-2}=2).

Exercise 2.

Solve the equation (sqrt{2x-8}=sqrt{3x-9}).

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Precalculus: the Mathematics of Numbers, Functions and Equations

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