 Hi, I'm Zor. Welcome to Unisor Education. Today's topic will be a couple of very small and very simple problems in the area of quadratic equations. Now, why these problems have to be addressed? First of all, there are some formulas which we touched during the lecture, the formulas for solutions of any quadratic equation. So, why do we need problems? Just have the formula and basically substitute all the coefficients in the formula and you've got the solutions. Well, yes, obviously. However, there are some reasons which really are much more important in this particular case. First of all, you can forget the formula and all of a sudden you're facing with a necessity to solve some quadratic equation and you forgot the formula. Well, that's not such a big deal because you always have to remember how the whole formula was derived and that's exactly why I want to solve a couple of problems because the way how I will solve these problems is exactly the way how we derived the formula during the lecture. So, it's easy to forget the formula. However, the approach you probably should always keep in mind and again in case you forgot the formula, you can always use this approach. At the same time, it's still useful to approach certain problems not from the formula basis but just from the logical thinking. How would you address this particular problem if you don't know any formulas? So, it's kind of a teaching problem, I would say. Illustrational for the approach we used during the derivation of the formula itself. So, I have three different equations here and first of all I would like to say that we are going to solve these equations in the area of real numbers. For complex numbers we will have different set of problems. So, number one, the domain is real numbers. Number two, I have different equations here. I have prepared three different equations. One will have two different solutions. Another will have only one and the third one will have none. But we will come to this gradually as it comes. So, number one is 2x square minus 5x minus 3 equals to 0. So, I don't know any formulas. I would like to derive some solutions of this particular equation purely logically, more or less the same way as I derived in general case, this general coefficients during the lecture. And that's exactly what I would like you to try. The approach which I will use, I will try to represent this as a square of something equal to something. For instance, basically x minus a, for instance, square equals to d. If I will be able to represent it in this particular form, then I can always derive x minus v, just square root of b and then x. So, this is also quadratic equation. So, if I will be able to derive from this, this with certain a and b, then my solution will be much easier. So, that's the general approach. Simplify your equation to a form which is easily to solve. In this case, in case of quadratic equation, this is basically the most simple and easily solvable type of quadratic equation. So, you have to transform this into this and then the problem will be almost solved. So, that's the approach. I would like you to think about this, press the pause button, try to do it yourself and I will basically do exactly this with the word right now. So, I would like to have something like x minus a square. Well, first of all, the coefficient with x square is one and this is not. This is two in this particular case. So, to represent this in this form, first of all, I have to divide everything by this coefficient with x square to have it one. Okay? So, my first transformation will be, I will divide the whole equation, which is invariant transformation by two and as a result, I will have x square minus five second x minus three second equals to zero. So, this is invariant transformation. I divided both sides of the equation by two. No solutions are lost or gained. Everything is fine. So, x square I already have, right? Now, if this is x square, then this is obviously this and then I will worry about all the three coefficients without any unknown variables x. So, if minus five second is minus two a, what does it mean? Well, it means that eight is equal to, I multiply both sides by minus one. That's invariant transformation and then divide by two. So, it's five four. So, I've got my eight. If I will change this into this, I subtract b from both sides. I will transfer b to the left and put it equal to zero. Then obviously a square minus b is minus three second, right? Now, I know that a is five fourth. So, what's the d? Well, if a square minus b is minus three second and a is five fourth, then obviously I will have five fourth square, which is 25 sixteenths minus b is equal to minus three second. So, what's the b out of it? Well, multiply everything by 16. So, it will have 25 minus 16 b equals to minus, that would be 24. So, 16 b, I add 16 b to both sides. So, 16 b will go on the right and then I will add 24 to both sides. 24 will go on the left. So, I will have 25 plus 24, 49 equals to 16 b. So, b is equal to 49 sixteenths. So, as you see, we have transformed our equation, original equation, first to this form and then using these a and b, we transform it into this form. So, I can say that my equation right now looks like x minus a, a is five fourth square, equals to b, which is 49 sixteenths. This and this are exactly the same. If somebody wants to make sure that that's really the case, let's just think about this. This is x minus five second x plus 25 sixteenths. This is 49 sixteenths. So, 25 sixteenths and 49 sixteenths, that's 49 minus 25 24. So, it's minus 24 sixteenths reduced by 8, 3 seconds. Yes, that's exactly what it is. We did not make any mistakes. All right. So, original equation, we have reduced this form, which is very easy to solve because now we can use the square root for both sides. This is not an invariant transformation that we will be very careful and we will use the absolute value of x minus five fourth. The positive part of it is equal to square root of 49 sixteenths is 7 fourths of this, right? Because 7 square is 49, 4 square is 16. Great. So, if this is a final equation, it means that either x minus five fourth is equal to 7 fourths. That's one solution. And another solution is 7 x minus five fourths is equal to minus 7 fourths. That's absolutely where it means. So, the positive, well, from this x is equal to 5 fourths. We add 5 fourths to both sides. We will have 12 fourths reduced by 4. That's 3. Now, 5 fourths on the right side minus 7. It will be 2 fourths. So, it will be 1 half, minus 1 half. Just in case, let's check. x is equal to 3 square will be 9 times 2 will be 18 minus 5 times 315. 18 minus 15 is 3 minus 3 0. Good enough. Minus 1 half. Square will be 1 fourth. So, it will be 2 fourths, which is 1 half. Minus 5 halves. Now, 5 halves is 1 half minus 5 half is minus 4 half. Now, minus, I think I made a mistake somewhere. Did I? 1 half is 1 fourth. All right. Let's do it this way. 2 times 1 fourth minus, that's plus 5 second minus 3. Minus 3. Now, plus because minus 5 and minus 1 half, it will be plus. Okay. Now, this is 2 fourths, which is 1 half plus 2 halves minus 3. This is 6 halves minus 3, which is 3 minus 3 0. Everything is fine. Okay. So, checking is okay. Both are solutions. And as you see, using this transformation into this form allows us to very easily solve the problem. Now, we will continue doing exactly the same thing with the other couple of problems, which I have. Okay. Next problem. Next problem is this. 4 x square plus 12 x plus 9 equals to 0. And again, my purpose is to reduce it to the form x minus a square equals d. Let's see what happens. Well, again, first of all, 4 really stands in the way because I would like that to be x square minus 2 a x plus a square minus d equals to 0. You understand I subtracted d from both sides, opened parenthesis and subtracted d. And now, 4 actually stands in the way. So, I have to divide the whole equation by 4, which is very interesting. No problem to that. And that would be x square plus 12 over 4. That's 3 x plus 9 fourth. So, my x square matches this one. So, what should be my a if 2 minus 2 a is equal to 3? So, my a is equal to minus 3 second. And a square minus b should be 9 fourth. Considering a is 3 second, a square minus b is 3 second square will be 9 fourth minus b. And that should be equal to 9 fourth. So, b is equal to 0 from here. So, my equation actually becomes b is equal to 0 and a is equal to minus 3 second. x minus 3 over minus 3 over 2, it will be plus 3 over 2 square is equal to b, which is 0. Let's check just in case that this is the same as this. x square plus 2 times x times 3 over 2, that's 3 x, 3 x plus the square of this one, which is 9 fourth. Everything is great. So, what's interesting about this particular example, this particular problem is that on the right I have 0. When I use the square root in both cases, I will have an absolute value of x plus 3 over 2 is equal to 0. Now, if absolute value of something is equal to 0, it's not like positive or minus or negative. There's only one 0. Last time we had something like, I don't remember, 7 over 2 or something like this. So, it was plus 7 over 2 and minus 7 over 2. In this case, 0 plus 0 or minus 0 is exactly the same thing. So, this is example when both solutions, I use the word both because usually quadratic equations have two solutions. Both solutions are the same. So, it's a double solution, so to speak. And the solution is obviously x equals to minus 3 over 2 to make it 0. Again, let's check it out. Just in case, 3 over 2 square will be 9 over 4 times 4 minus 12 times 3 is 36 over 2 plus 9. Now, this is 9. This is minus 36 over 2, 18 and plus 9, 0. Everything is fine. So, that's an example of an equation which has only one solution. Well, we use the term double solution if you wish. But anyway, it's only one number. That's example number 3. And number 4, sorry, that was number 2 and this is number 3. Okay, the last one. 3x square minus 18x plus 30 equals to 0. Good moment to press the pause button and do it yourself right now. Now, I will do exactly the same thing as before. I want to transform it into this particular form. Or if you wish, x square minus 2ax plus x square minus b is equal to 0. That's the form I would like to transform it for. Okay, now, this is x square, this is 3x square. So, I divided everything by 3. That's number 1, obviously. So, I have x square minus 6x plus 10 equals 0. Great. Now, if minus 2x is minus 6x, then a is equal to 3, right? a is equal to 3. Minus 2 times 3 times x, that's exactly right. Now, so a square minus b is supposed to be 10, right? Now, a is 3. So, it's 9 minus b equals 10, from which b is equal to minus 1. And that's very important. b is equal to minus 1. You see, b is on the right. On the left, we have a square of something, a real number. Don't forget, we're solving in the real numbers right now. The domain of the solutions is the set of all real numbers. Obviously, no real numbers square will give minus 1. That's the complex numbers, where we will have the number i, which will give this. But that's not what we're considering right now. Right now, we're talking only about real numbers. There is no real number square of which is equal to minus 1. So, what does it mean? It means there is no solution. So, among the real numbers, this particular equation has no solutions. And that's the final point, which I wanted to make with this problem number 1, which is actually three different little exercises. And again, the most important part so far was to transform our original equation into this form, which is very easily solvable, just because we can have the square root of both sides, if possible, if it's not negative on the right. Okay, that basically concludes the algebraic part of these three equations. Now, I would like to illustrate everything geometrically. If I have an equation, then whatever is on the left, which is a polynomial of the second degree, power of 2, it's represented by some kind of parabola. Now, the horns of this parabola are directed either upward or downward, depending on the coefficient and if I have something like ax square plus vx plus c is equal to 0. If this is my original equation, then this coefficient determines what exactly the direction of the horns of the parabola are. If it's positive, then it's upward. If it's negative, it's downwards. So that's number 1. Number 2, solutions to this equation is when this parabola is crossing the y equals 0 axis, which is this horizontal axis. So this parabola has two solutions. This parabola has no solutions because it doesn't cross the x axis and this parabola, which just touches this particular x axis, has one solution which can be called double. So my original three equations which I had are actually illustrations to all these three cases. One case when parabola doesn't touch it, another is when there are two solutions and yet another one when there is only one solution, but it's downward. Now, let's just make it more concrete in each of those cases. So I will try to draw a graph and we will see how it actually illustrates. So again, let's start from the beginning. 2x square minus 5x minus 3 is equal to 0. I divided this thing by 2x square minus 5 to 2x minus 3 is equal to 0 and represented this as x minus 5 over 4 square, because that would give me 5 over 2x square minus 5 over 2x and then I have 25 over 16. So I have to subtract 25 over 16 and minus 3 and that would be my equation. So just let me combine these together. 3 is 25 16 and minus 3. So what were the roots of that? I don't remember. I thought it was plus. Let me just check it again. So I have x minus a square is equal to b, right? So it's x square minus 2x to ax plus a square minus b is equal to 0. So this is covered by this. So a is equal to 5 fourths. So 25 over 16, here it is. Minus b is equal to 0, sorry, is equal to minus 3 is equal to minus 3. Right? Minus 3. So b is equal to, b goes to the right, 3 goes to the left. It's 48 minus 25 16. Am I right? Something is wrong. No, plus. Plus is no good. This is 5 fourths. Let me check. But oh, I know why. That's my mistake. It's not 3. It's 3 seconds. I divided by 2. See, that's my mistake. Okay, fine. So let's just change this a little bit. I see some unusual numbers which I didn't really plan for. Okay, now it's different. So now it's ix minus 5 fourths square plus, well, actually minus 25 16 because this is plus 25 16. If I open the parenthesis and minus 3 second is equal to 0. Okay, now that's better. Because now this is instead of 3 second, I will put, multiply by 8, it will be 24 16. Right? Now minus 25 and minus 24 16 is 49 16. So I'll change this to minus 49 16. Now that's a familiar number. I remember that. Okay, now I can draw a graph of this particular polynomial. Now, if you remember, if I take the graph function y is equal to x square, that would be a parabola which goes like this. Almost perfect. Now, next is let's draw the graph x minus 5 fourths square. Now, if you remember what happens with the graph, if I substitute instead of x, x minus something, well, the whole graph shifts to the right by this something. For obvious reasons, so in this case, if this is 5 fourths, the parabola will be like this, new parabola. Because for the value of x is equal to 5 fourths, this graph is exactly the same as this one when x is equal to 0. And for the x is equal to 5 fourths plus anything, it's the same as this graph when x is equal to that anything. So the whole thing is actually shifted to the right. This graph is shifted by 5 fourths to the right from the original y is equal to x square. So that's covering that we were talking about this part. Now we have to subtract something from the graph. Now subtract means the whole graph shifts down by, in this case, since it's minus it's down, so it's 49 sixteenths. So if this is minus 49 sixteenths, the whole graph will shift down to this position and still its horns will be upwards. And wherever the roots of this equation are, and the roots are, let's do it again, it's x minus 5 fourth absolute value is equal to square root of, which is 7 fourths. Now, which means that x minus 5 fourth is equal to plus or minus 7 fourths. If it's plus, then it will be 12 over 4, it will be x is equal to 3, and if it's minus it will be 2 over minus, okay, minus 1 half. So this point is minus 1 half, and this point is 3. Well, the scale is obviously not very good, but in any case, this is the final parabola and crossing of this parabola with the x axis, which is 3 and minus 1 half, are exactly the solutions. So this is a graphical representation of the same thing. Now let's continue with the two other examples. For example, number two, I hope I will not make an original mistake with number two. So we have 4x square minus plus 12x plus 9 is equal to 0. So first I divided it by 4, so it's x square plus 3x plus 9 fourth is equal to 0. Now this is x plus 3 over 2 square, just to get 3 here, x square plus x times 3 seconds times 2, that's 3x, and so this is already 9 fourths. So basically my 9 fourths is already covered. That's the same thing. So this is parabola again, which is shifted by 3 over 2, but to the left in this case, because this is plus sign. So geometrically it looks like this on the graph. If this is minus 3 over 2, it will be parabola, which looks like this. And again crossing with the x axis is only one point, and this is the point minus 3 over 2. Finally, my third example where if you remember we did not have any real solutions is 3x square minus 18x plus 30 equals to 0. First we divided by 3, that was easier, minus 6x plus 10 is equal to 0, this is x minus 3 square. Now that would be 9, right? So I have to minus 9 and plus 10 is equal to 0. So this is minus 1, I mean plus 1. Okay, let's draw the graph again. We have the original location of our parabola, shifted first to the right by 3, that's my this one, but now we have to move it up by one. So the whole graph will look like this. This particular parabola as you see has no crossings with an x axis, and that means there are no real solutions. Well, that basically concludes all three examples which I wanted to talk about, so let me just summarize very briefly. What's most important in all of these cases was the illustration that you don't really have to remember relatively complex formula for quadratic equation, for solutions to quadratic equation. If you have any quadratic equation, your purpose is to bring it to this form, from which you can obviously very easily conclude that x minus a absolute value is equal to square root of b, which does have real solutions if e is not negative, so x minus a is equal to square root of b or minus x minus a is equal to square root of b, or if you wish I can multiply by minus 1 both parts and f minus here, or if you wish to do it even more shortly, that's how sometimes it's expressed. Both solutions are expressed in one formula, but using plus minus. From which obviously we have x is equal to a plus minus square root of b, by adding a to both sides. So this is a solution if you transform your equation into this format, then this is the obvious solution. And that's probably the most important part which you have to remember from these exercises. Just simplify your equation to this form and then the solution will be very natural and easy. And from the graphical representation, again remember that parabola is supposed to cross the x-axis in either two points, then there are two solutions, that's when the b is positive. One point if b is equal to zero, and that would be the point x is equal to a solution and one and only. And then there are no solutions if b is negative, because the square of the real number cannot be negative. There are no x's. Well, that's it for today, good luck. And don't forget, check the unisor.com where there are many other lectures and problems and exercises available. Thank you very much.