Skip to 0 minutes and 11 seconds Hello. In week five, we’re dealing with inequalities. It is quite a delicate topic even for us. In this first step in practice of week five, we are studying the general principles. And more precisely, we are dealing with the sign of a quotient. In exercise one, we first find the domain of inequality, which is the set of x, such that the denominator x minus 2 does not vanish. That is, the set of real numbers except 2. Notice also that for x in the domain, the sign of the quotient x plus 3, times x plus 1, divided by x minus 2 is exactly the sign of the product of these three terms.

Skip to 1 minute and 3 seconds x plus 3 times x plus 1, times x minus 2. And so we use the product rule. So we are going to construct the table with the signs of the three terms– x plus 3, x plus 1, and x minus 2. And then we study the sign of the product. Now the values that vanish x plus 3 is minus 3, x plus 1 vanishes in minus 1 and x minus 2 vanishes in 2. So we’ve got three important values– minus 3, minus 1, and 2. So let us construct the table. And we shall insert here the values minus 3, and minus 1, and 2.

Skip to 1 minute and 51 seconds Here we study the sign of x plus 1, which is 0 at minus 1– positive on the right hand side and negative on the left hand side. And then here we study the sign of x plus 3. And so which vanishes at minus 3. It is positive here and negative there. And then finally the sign of x minus 2, which is 0 in 2– positive on the right hand side and negative elsewhere. Here, the sign of the quotient, x plus 3, times x plus 1, divided by x minus 2. Let us recall that this quotient is not defined at 2, so we put the double line– double vertical line just to remember.

Skip to 2 minutes and 42 seconds And the other signs are exactly the sign– the sign of the quotient is the sign of the product. And so here we have a negative sign, here we have 0, here we have a positive sign, here we get 0 again. Here we have a negative sign and here a positive sign. So in order to have this quotient less or equal than 0, we have to stay on this interval or in this interval, except 2. So the solution is the union of the two sets, of the two intervals minus infinity, up to minus 3. And we take minus 3, union, well, the interval from minus 1 to 2, but we do not take 2. Now, how about exercise 2?

Skip to 3 minutes and 41 seconds Well, we have a similar inequality with the difference that instead of x, we’ve got square root of t. So the domain now– D prime– is the set of t– well, such that square root of t is defined. And so we have to impose t greater equal than 0, and also square root of t must be different from 2. So the square root of t, different from 2. That is 0 plus infinity, except t equal 4.

Skip to 4 minutes and 25 seconds Now, let us put x equal to square root of t, and we already solved the inequality with x. So we impose that x must belong to minus infinity minus 3, union, minus 1, 2. That is, the square root of t belongs to this union of two intervals. Now, square root of t is positive. So actually, this is equivalent to impose that square root of t belongs to the interval 0, 2.

Skip to 5 minutes and 10 seconds And this is equivalent to t belonging to the interval 0, 4. Every element of this interval belongs to the domain of the inequality, and so the solution here is the interval 0, 4.

Skip to 5 minutes and 31 seconds So see you in the next step.

# General principles 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 \[\dfrac{(x+1)(x+3)}{x-2}\le 0.\]

### Exercise 2.

Solve \[\dfrac{(\sqrt t+1)(\sqrt t+3)}{\sqrt t-2}\le 0.\]

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