 Hi, I'm Zor. Welcome to a new Zor education. I would like to continue talking about inequalities. I'm pretty sure you're bored with linear inequalities. They're kind of simple and they don't challenge your mind at all. So, I want to spend some time with a little bit less trivial inequality problems. So, there is not too much to talk about some kind of a theory about how to solve general inequalities. The only thing, just a couple of common sense, if you wish, comments about how to solve inequalities. First of all, you really have to look at the inequality as it is given to you and the restrictions, this particular inequality, imposes on whatever arguments participate in it. If you have, for instance, you have unknown in denominator. Obviously, you have to check a special case when this denominator equals to zero because you cannot divide by zero. So, these values must be excluded from whatever further solution you have received. Another thing which is very important is, if you have some non-invariant transformations of inequality, you really have to be very careful about this. For instance, you're reducing your fraction by some multiplier in the denominator and numerator, and this particular multiplier has an unknown variable. That's, again, another thing which you have to really check very thoroughly when this particular multiplier is equal to zero. So, in any case, allowed value, non-invariant transformations, these are to be paid attention to. Now, also, checking. Well, inequalities usually produce infinite number of solutions, so you can't really check all of them. However, a spot check of certain more or less crucial points, like zero, one, minus one, or something like this, it will also help in any case. Alright, so, without further ado, let me just go to problems. I have three problems to solve during this lecture, and I'll just present these one by one, and the more problems you solve, the more you're equipped with solving future problems. And again, there is no recipe. It's just particular cases which will somehow grow into your general understanding of approaches, how to approach it. Okay, number one, 2x plus 7 times 3x minus 5 divided by 4x plus 9 greater or equal than zero. So, we are looking for the result of certain multiplication and division of simple expressions, because each one of them is linear, and we are looking for the result of this to be non-negative. Let's just think about this from the following perspective. When a particular result of multiplication and division can be non-negative. Well, number one, when everything is positive or non-negative. Then, when two of participants are negative and one is positive, because negative times negative gives you positive gains. So, if you have two participants among these three, for instance, positive, then the result will be positive. If it's more than one to three components of multiplying and dividing, if it's like a hundred different components, then if you have an even number of negative values which we are multiplying by each other, then the result will be positive again. So, we really have to check when particular components of this multiplication or division of some algebraic expressions is changing signs. Now, the best way to approach it, the same by the way as we were using when talking about absolute value, is to find out all points where each component equals to zero. So, we will have this line, an American line, this is zero, and now we will mark on this line points where these guys are equal to zero. Okay, this one is equal to zero when x is equal to minus seven seconds, right? So, this is minus seven over two. Now, this one is five over three, which is something here. And this one is minus nine-fourths. Nine-fourths is somewhere here. Seven seconds is more than three by absolute value, right? Nine-fourths is just a little bit over two. That's why it's closer to zero. Both are negative. So, these are the points where each corresponding member is zero. So, let's just examine separately the intervals of all the real numbers which these three points are dividing. Now, what if it's greater than five-third? Well, then this will be positive, this will be positive, this will be positive. So, everything will be positive. Let's put plus. Now, whenever the point moves from five-third to minus nine-fourths, one particular component, which is this one, changing the sign. Others do not, right? So, this is a negative area. When you move forward, I mean really to the left, we are moving to the left. One more component becomes negative. So, this one and this one becomes negative, if x is in this particular area. So, negative and negative gives us positive. So, the whole expression will be positive. And then again, when the third one, which is this one, changing the sign, the whole expression will be negative again. So, we will have an alternating sign of the expression as the argument moves from left to right or right to left, minus plus, minus plus, or plus, minus, plus, minus. They are always alternating because after crossing each point, only one particular expression out of these three in this particular case changes the sign. So, once we have established this, we can very easily say that this will be expression greater than or equal to zero when the argument would be in this area and in this area. The only little problem is, you see, this is denominator, which means we should really avoid the point when it's equal to zero. So, these are okay, but this point, minus nine-fourths, must not be included. But everything else is fine. So, our total solution is x should be in this interval and pay attention to this strictly less. This is less or equal. This is strictly. Meaning equal to minus nine-fourths because the denominator will be equal to zero. And x is greater than five-thirds. So, the combination of these two areas, so it's union of these two sets. This set is unionized with this set and that gives you a total solution. Okay, now, is there any way we can check it, just spot check if you wish? Well, this is a little bit to the left from two and this is, I'm sorry, this is second, not fourth. And this is less than minus three. So, something like minus three should be in between, right? So, let's just check if minus three. Now, this is minus six plus seven, this is one, which means it's positive. If x is equal to minus three, this is negative. It will be minus fourteen and this one will be minus twelve plus and minus three, also negative. So, positive, negative, negative, so result will be positive. So, it's okay. Now, let's go to the right. Let's go to zero. Now, it's not part of our solution and let's check what happens. If x is equal to zero, it's seven times minus five, which is minus thirty-five divided by positive nine, so it's negative. It doesn't fit. So, basically, we are spot checking this thing and that's the result of the solution. So, this is the union of these two areas, e is a solution. The end. So, again, let me just summarize it as an approach. If you have a combination of different expressions which are multiplied or divided by each other and the result should be either greater or greater or equal or less than zero, then you can always examine the sign of each particular expression, check where this sign is changing. So, the point where it's zero and then on both sides of zero. And that would give you the right breaking of the entire set of values of x into separate intervals in which each of these has a specific sign. So, you can always examine the sign of the result. Now, the second one is very similar to the first. However, there are a couple of interesting details. Now, this is exactly the same thing as before, but numerator is multiplied by x square and denominator is multiplied by x minus two to the fourth. Well, this looks much scarier, right? Let's just again think about it. Don't forget that x square is always non-negative, right? So, the fact that we have multiplied our original inequality by x square in the numerator doesn't really add much to this particular inequality, except maybe one case when x is equal to zero. Because in this case, if x is equal to zero, this will be the equality, right? And the equality is good for us because we are looking for greater or equal. So, regardless of the sign of this, if x is equal to zero, then the equality would be held and that's why it would be a solution. Now, in this case, we are dividing by something to the fourth degree. Now, obviously, we should not divide by zero, so that restricts our set of allowed values. Now, it's also the fourth degree, which means it's positive or zero. So, if it's positive, it doesn't really change the sign of this thing, but if it's zero, it actually restricts the value. So, my general answer to this particular problem is take whatever the solution you had without these two components. And if you remember, this is 5 third minus 9 fourth and minus 7 second. So, if you remember, we had this area, but I will put a little arrow here, which means minus 9 fourth is not included, and this area, including and here and including and here, right? So, that's the original one, but now we have to add and subtract something. If x is equal to zero, which is somewhere here, it's a solution. If x is equal to 2, which is somewhere here, this is not a solution. So, the resulting solution would be, again, from minus 7 second to minus 9 fourths, not including the end. Here I have equal sign included, here not. Then, x is equal to zero. This is this point. We should add it because then the whole thing becomes zero and that's an equal sign. Then, the original solution was greater or equal than 5 third. That's fine, but I should not include point 2 and everything greater than 2 is okay as well. So, it's less than 2 or greater, but not equal. So, that's how I exclude. So, these four areas, this area actually contains only one point. These are real intervals. Unionized together constitute the solution to this problem. What's the kind of main point? You have multiplied something which you already know what it is by an expression with unknown inside and that basically changes the picture. And you divide by something which contains unknown. You have to carefully examine when this is equal to zero, as much as we have examined this equal to zero, we excluded the point minus 9 fourths. So, we should really exclude the point 2 as well. But in this case, this point belongs to the edge of the interval. In this case, it's in the middle of the interval, but whatever. So, that's the second problem. And the third one, which I wanted to talk about was square root of x plus 10 greater than 2x plus 5. Alright. Whenever we are talking about square roots, we always talk about arithmetic value, arithmetic value, which means this is a positive value from a non-negative value from a non-negative argument. And it's mandatory to have non-negative here. So, we have already a restriction that x should be greater or equal than minus 10 for this expression to be non-negative. And the square root, the arithmetic value of square root is also non-negative whatever it is. Now, what it means actually, that whenever this part is negative, everything already correct, because this is positive or zero. And if this is negative, then basically all the values which fit this when this is negative and this basically exists are solutions. So, for these values now, when is this particular expression is negative? When x is less than minus 510, right? Am I right? This is equal to zero when x is equal to minus 5 seconds. And if x is less than this expression is negative. Now, if this particular expression is negative, it means that all values are good actually. They are solutions. So, everything from minus 10 to minus 5 seconds, minus 10x minus 5 seconds, are solutions we have already established in just examining the sign of this and the sign of this. Because in this case this would be a positive or zero, which means the whole thing would be non-negative. And this thing would be negative. So, this is non-negative and this is negative and obviously there is a greater relationship between them. So, these are all solutions. Now, let's consider a different case when x is greater or equal than minus 5 seconds. Now, what happens in this case? Well, in this case this expression is non-negative, positive or zero, right? So, this is non-negative and this is non-negative, which means that we can actually in this particular case just square both sides. Because both sides are non-negative and if we square them the relationship between them remains the same. Now, this is a non-invariant transformation, generally speaking, squaring both sides of the equation. However, in case when both sides are non-negative, there is no problem. If they were non-negative, after that they will be non-negative or vice versa. So, it becomes invariant basically when we know that both expressions are in this case non-negative. By the way, this implies this obviously. So, this second case is a pretty universal restriction on our x and let's consider what happens in this case. Alright, so we square both sides and we will have x plus 10 greater than 4x squared plus 20x plus 25. Or 4x squared plus 19x plus 15 less than zero, right? This side x, 20x minus x on both sides, I have 19. 10, 25 minus 10 on both sides will be 15. And this sign relative to this, I put it on the left now, it's less than zero. Okay, now this is a parabola, right? Now, parabola has two horns, in this case directed vertically upwards. Now, if parabola has two points where it crosses the x-axis, then this area is where it's negative, right? That's what we're looking for. So, what we have to do is we have to find out if this particular quadratic expression has roots, has solutions of the equality when it's equal to zero. And then, if we want to know whether it's where exactly it's less than zero, we should really take this interval in between these two roots of this quadratic polynomial. Well, let's check it out. Does this particular equation have any roots, have any solutions? Well, let's try minus 19 plus minus 19 squared is 361 minus 4 times 1560 times 4 to 40 divided by 8. This is, now, this is positive, so there is a root, so there are two solutions. Now, this is 121, so square root of 121 is 11. So, basically this is minus 19 plus minus 11 divided by 8. So, one solution is equal to when it's plus, it's minus 8 minus 1. And another solution is minus 3815 fourth, minus 1544. So, what I might add to this whole picture that this particular inequality, this particular inequality is true in this particular interval. So, this is minus, almost minus 4. So, it's from minus 15 fourths x less than minus 1. That's what it is. However, okay, you don't need this anymore. This is minus 15 fourths, this is minus 5 seconds, and this is minus 1. So, here's what's interesting. Solution to this particular inequality is from minus 15 fourths to minus 1. However, we only have to deal with unknowns in this particular interval, which means only from here, what does it mean? It means that solutions are only in this area, from minus 5 seconds to minus 1. Minus 1 is not included, by the way, and minus 5 seconds is included. So, I have to add to this particular interval another union with minus 5 seconds x less than minus 1. So, the combination of these, and this we can actually easily unionize together, because it's from minus 10 to minus 5 seconds, not including minus 5 seconds, but now from minus 2 seconds including to minus 1. So, the result is minus 10 less than or equal x to minus 1. So, this is the final solution, the union of these two areas. So, this is an equation which has this particular solution. Well, that's it for today. It's a very simple, but still a little bit less trivial inequalities of general type. I will present probably another set of a few problems, slightly more challenging maybe, but still non-trivial. That's what we actually are looking for. Well, just think about what kind of approaches I was using. You remember the first couple of problems where with multiplication and division, and you can basically divide the whole interval into areas where they're changing signs. In the second case, what's important was also consider the allowed values for unknown, before you really do any transformations, invariant or non-invariant. So, that's it for this particular lecture. I hope you enjoyed it. Well, wait for yet another one with equate inequalities, a little bit less trivial. Thank you. Good luck.