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5.12

## Precalculus

Skip to 0 minutes and 11 seconds Hello, and welcome back to a step in practice. Today, we are dealing with inequalities that involve radicals. It is quite a difficult topic. It’s not a coincidence, probably, that I’ve had to make this video about 15 times. I hope this is the last. Now, in exercise 1, the inequality is square root of 9 x squared plus 18x, strictly greater than 3x plus 2. First of all, we have to study the domain of the inequality. The domain here is the set of x, such that the square root of 9 x squared plus 18x makes sense. That is, the set of x, such that 9 x squared plus 18x is greater or equal than 0.

Skip to 1 minute and 9 seconds Now, 9 x squared plus 18x is a second degree polynomial whose roots are 0 and minus 2. And therefore, since 9 is positive, the polynomial is positive before the first root and after the second root. So the domain is the union of the intervals minus infinity minus 2 union 0 plus infinity. We have to take care of this set because at the end of our calculations, we have to remind that x must belong to this set.

Skip to 1 minute and 51 seconds Now we solve the equation. And we use the method that Francis showed us in the previous step. The inequality square root of 9 x squared plus 18x greater than 3x plus 2, when x is in the domain, is equivalent, well, to the square of the equation, that is, 9 x squared plus 18x greater than 3x plus 2 to the square or 3x plus 2 is strictly negative. Now, let me point out two things. Most of you consider only the first inequality and forget the second. This is a pity because here we have plenty of solutions, all the x of the domain that are less than minus 2/3.

Skip to 3 minutes and 3 seconds The second point is that while here we have a union of solutions of the two inequalities, there is the or, so we take the union of the two solutions. In order to remind these, let us remark that here we require the square root of something to be bigger than. So the square root to be big. And if you have two sets, the bigger between their intersection, or their union, is their union. So we take the union of the two solutions. This is a way to remind it. Now, the second inequality immediately gives us x strictly less than minus 2/3.

Skip to 3 minutes and 56 seconds Let us study the squared inequality. Now, 9 x squared plus 18x strictly greater than 9 x squared plus 12x plus 4, while the 9 goes away, 9 x squared goes away, so this is equivalent to 6x strictly greater than 4, that is, x strictly greater than 2/3. So the solution, the union of the two solutions is x strictly less than minus 2/3, or x strictly greater than 2/3. But now we have the compatibility with the domain.

Skip to 4 minutes and 42 seconds We have to take just the x that belong to the domain. So while we cannot take x after minus– negative x after minus 2, so the solution is minus infinity minus 2, and then here we can take every x greater than 2/3 because the domain, the positive elements belong to the domain. So union 2/3 plus infinity. And so this is the solution to exercise 1.

Skip to 5 minutes and 23 seconds In exercise 2, we are dealing with the inequality with radicals in the reversed sense, namely, square root of x squared plus 1 less or equal than x/2 minus 1. But the domain here is the set of real numbers because x squared plus 1 is positive for every x. Now, this inequality is equivalent to, well, the squared inequality, x squared plus 1 less or equal than x/2 minus 1 to the square AND well, x/2 minus 1 greater or equal than 0. And this is a system of inequalities. Let me point out here that we have the word “AND”, so we have to– we are going to take the intersections of the two inequalities.

Skip to 6 minutes and 38 seconds Again, a way to remind this is that here we want the square root to be small, and if you have two sets, the smallest between their union and their intersection is their intersection, so we take the intersection of the two inequalities Also, let me point out here, and also in the previous exercise, that here we have a large inequality and we found a large inequality here. Here we are restricting equality. And it’s not by chance that we find a strict inequality here.

Skip to 7 minutes and 20 seconds Let us solve this system of inequalities. Well – the second one gives us the constraint that x must be greater or equal than 2. Let us write this in red because we must take this into account at the end of our calculations. And let us consider now the squared inequality. So the squared inequality states that x squared plus 1 is less or equal than x squared over 4 minus x plus 1. Let us multiply these by 4, the 1 goes away. And so this is equivalent to 4x squared less or equal than x squared minus 4x. That is 3 x squared plus 4x less or equal than 0.

Skip to 8 minutes and 26 seconds Let us put x in factor, and this means x times 3x plus 4 less or equal than 0. Now, the roots of the second-degree polynomial are minus 4/3 and 0. So the polynomial of second degree will be negative between the two roots. This is because the coefficient of x squared is positive. So the solution here is x belonging to the interval minus 4/3 and 0. Now, let us look the compatibility.

Skip to 9 minutes and 14 seconds We want x belonging to the domain, x greater or equal than 2, and x belonging to minus 4/3, 0. Well, there are no x that satisfy these conditions, so the set of solutions is the empty set. And this ends exercise 2. The solution to exercise 3 is proposed in the PDF that you will find at the end of the page. So see you in the next step.

# Inequalities with 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 following inequality: $\sqrt{9x^2+18x} > 3x+2$

### Exercise 2.

Solve the following inequality: $\sqrt{x^2+1} \leq \frac{x}{2} -1$

### Exercise 3 [Solved only in the PDF file].

Solve the following inequality: $\sqrt[3]{x-1}>\sqrt[3]{5-x}.$ (Recall that in this course the cubic root is defined just on $$[0,+\infty[$$)