Skip to 0 minutes and 11 seconds 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.

Skip to 1 minute and 9 seconds 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.

Skip to 1 minute and 58 seconds 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.

Skip to 2 minutes and 45 seconds 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?

Skip to 3 minutes and 55 seconds 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.

Skip to 4 minutes and 41 seconds 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.

Skip to 5 minutes and 19 seconds 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.

Skip to 6 minutes and 1 second 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.

# Equations involving several radicals in practice

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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