 I'm Zord. Welcome to Unizord Education. Today's topic will be quadratic equations. Well, the last thing I would like to do talking about quadratic equations and probably about any kind of equations or some theoretical topic is just to give you a formula. Okay, this is quadratic equation. This is the formula for its solutions. Just use it. Well primarily because this is definitely not the purpose of these lectures. The purpose is to basically explain how the real thinking is going on behind it and what leads to these equations and solutions and the methodology, etc. So that's why I will try to do it by introducing certain different categories of quadratic equations more and more complex and come up with solutions just as if I will try to solve it myself from the first time without knowing any kind of formula. So I don't suppose you know any formula and let me just try to derive all these formulas and solutions in this particular case as we go. So first of all what is a quadratic equation? Anything which has a form of this where a is not equal to 0 represents a quadratic equation where x is unknown, a, b, and c are known coefficients. Well first of all when we talk about equations we have to think about where exactly we are trying to solve this particular equation. What's the domain of our unknown x in this particular case? Well there are two traditional domains for solving equations and quadratic in particular. The first one being domain of all real numbers and the second one is complex numbers. So today we will talk only about real numbers. So a, b, and c are real numbers and x is a solution which we are looking within the domain of real numbers only. Alright so that's the definition of what's the quadratic equation. Well obviously the solution is any real number x which being substituted to this expression will result in 0. Okay so now we are facing the problem. Let's solve this particular solution. Let's find all the possible values of x real values which represent the solution to this equation. And quite frankly if I first time I look at this equation I just have no idea how to approach it. I'm just thinking that well there are some simple quadratic equations which I know how to solve. For instance this is the simplest quadratic equation which being represented in this particular form would look like x square minus 1 is equal to 0 because this is an invariant transformation of this equation to this equation. We subtract 1 from both sides. Now in this particular case a is equal to 1, b is 0 and c is equal to minus 1. Okay so this is an example or this is an example of this particular equation and I can just guess how to solve this equation. Obviously it has two different solutions x is equal to 1 and x is equal to minus 1 or in abbreviated form we can have plus minus 1 which is the same thing. So these two and this one mean exactly the same thing. Alright so we know how to solve this equation. Well let's try to make a little bit more complicated example. What if my equation is not this but something like this. Can I solve this equation? Well I can actually try to do exactly the same thing as in the previous case. I use the invariant transformation at r to both sides. I will get this. Now if r is not negative then solution exists and one of them is x equals to square root of r and another is minus square root of r. By the way when we are talking about square root you always have to remember we are talking about a arithmetic value of the square root which means the positive number square of which will give r. So in case of r is equal to 25 square root is 5 not minus 5. So square root is 5 and when I am saying minus in front of it that's what minus 5 actually is. So x is equal to 5 and minus 5 are both solutions to x square equals to 25. So basically what I'm saying is that I can solve this equation or this equation more or less as easily. Well that's complicated a little bit more. I have this one. I can do exactly the same thing as before but I will use x plus s. I subtract r in both cases. Obviously in this case r is supposed to be less than 0 or equal to 0 or I can do just looks a little easier. It's silly to write it this way. Let's use it this way and now the requirement is r is greater or equal to 0. That's kind of easier right? So how to solve this particular equation? Well same way as before I can say that x minus s is equal to square root of r where x minus s is equal to minus square root of r as in the previous case. And applying invariant transformation of adding s in both cases I can have two solutions. One of them is x is equal to s plus square root of r and x is equal to s minus square root of r. Two solutions. Four s is equal to s plus minus square root of r. Alright you see we can solve very easily certain number of equations. Well one more complication and then we will go to the what if I have this? No big deal again from the very beginning I assume that a is not equal to 0 by the way the same as in this case that a times x minus s square is equal to r. Now since a is not equal to 0 divided by a is in invariant transformation so x minus s square is equal to r over a and everything else goes as before the only condition is r over a greater or equal to 0 from which follows that s x is equal to s plus square root of r over a and x is equal to s minus two solutions. So this is the most complicated equation which we don't really which we don't really have to think twice about how to solve. This on the other hand is slightly different but let's compare these two equations one of them which we know how to solve and another with them. Let me just put solutions to this equations here and s minus r over a. Okay these are two solutions to this equation and this we don't know. Well let's compare these two things maybe I can somehow transform preferably invariantly transform this into this. Let's open up this expression it's a times x square minus 2 x s plus s square right that's what x minus s square is minus r is equal to 0 multiplied a x square which looks like this by the way that's why I was using x minus 2 a s x a s x plus a s square minus r is equal to 0. Well you see it's really more or less very much like this one where this is b and this is c. This is the coefficient with x and this is c which is a free element. So if I will be able to find a s and r s and r in such a way that this expression is equal to b and this expression is equal to c then I can use this formula to get the final solution. So this formula expresses the solution to this equation in terms of s and r but if s and r are expressed as b and c then using the same formula just using b and c instead of s and r I will have the solution to this equation in terms of a b and c which is definitely what's required. So all I have to do is using these two equivalences find representation of s and r in terms of a b and c. Well let's do it it's really not very difficult. So from this equation from this equivalence of minus 2 a s is equal to b. I can immediately derive that s is equal to minus b over 2 a correct? Remember a is not equal to 0 so dividing by 0 is allowed so s is equal to b over minus 2 a. Now I can use it here to find r represented from this equation. So it's a times s square which is this which is b square over 4 a square right that's what s square is minus r is equal to 0. Now I don't need this and I have my expression of r by the way a can be reduced here but this is multiplication this is a square so r is equal to b square over 4 a. Now this this are representations of r and s in terms of a b. Wait a minute I'm sorry it's not equal to 0 it's equal to c. I made a little mistake here sorry about this. So r is equal to b square over 4 a minus c right r goes to this way c goes to this way. This is my expression. I was very much surprised that c did not participate in the formula. All right so now since my r and s are represented in terms of a d and c I can just substitute it here and I have a solution basically right so let's just use these two representations put it here and off we go. So the first solution is write these two formulas here s is equal to minus b over 2 a and r is equal to b square over 4 a minus c. Now I have some real estate to put my formulas. So using these expressions for r and s in terms of a d and c I will substitute it into this expression for x and I get and I have two solutions solution number one let's put number index number one and two so solution number one is equal to s plus the square root s is minus b over 2 a plus square root of r which is this. If you don't mind I will use the common denominator 4 a in this particular case it's b square minus multiply 4 a by c it's 4 a c divided by 4 a it's a little easier all right so this is r and I have to divide it by a so I will have b square minus 4 a c divided by 4 a times a it's 4 a square and the second solution x2 is equal to almost the same thing except it's not it's a minus here so these are my two solutions the only thing is I would like to simplify these formulas just a little bit you see this is 4 a square which is a full square of either 2 a or minus 2 a right 2 a square will be 4 a square minus 2 a square will always be also will be 4 a square so I don't know what sign of a actually has but you see I have two solutions here plus and minus so it doesn't really matter what sign of a is I can always replace this with leaving the square root only from the top part and changing this to a real square root which is 2 a since I have plus here and minus here it's still two solutions doesn't really matter what's the sign of a is if a is positive it will be a solution or if a is negative is also a solution which is the same thing right but now this is 2 a and this is 2 a so it makes sense just to have a common denominator it will be plus and this will be minus so these are two solutions to our original quadratic equation minus b plus square root of b square minus 4 e c and minus b minus square root of b square minus s and obviously the whole thing exists only if as we see this is another solution if you remember in terms of r and s we were always um putting r greater or equal to zero well basically that's the same thing but in terms of a b and c so this is a condition for quadratic equation to have two different or the same maybe if it's equal to zero square roots solutions to square to quadratic equations now this is a condition for an equation to be quadratic because if it's if it's a is equal to zero then it's not a quadratic equation and this is a condition of quadratic equation having solutions if it's a strictly greater than zero these are two different numbers obviously and it's two different solutions if it is equal to zero which is b square minus 4 a c is zero then we have two solutions which are really one in the same because it's plus zero or minus zero a is not equal to zero and so there is nothing wrong with having a in the denominator well as an exercise I would definitely ask you to substitute x1 and x2 both in this original equation and check that this is indeed solution so if you will substitute this square the whole thing times a plus whatever is necessary you will really get zero in this case as well as in this case now just to jump a little bit forward we will probably spend some time to solve in this equation in the domain of complex numbers in the domain of complex numbers we don't really need to to have this restriction because in the domain of complex numbers square root of any negative number exists like square root of minus one for instance it's i in terms in terms of complex numbers so what's interesting is the quadratic equation always has solutions in the area of complex numbers in the domain of real numbers it's only under this condition okay that's good and we will spend some time to analyze the geometrical interpretation of these solutions so let's go to geometry I think I will use again the same approach of starting with something simple and then going into a little bit more complicated cases so let me put these formulas on the side actually I don't think they will be needed anyway as well as this condition and let's just think about this particular equation well first of all I would like to to graph this left part of this equation as a function so if I have a function y equals two and in this case I will use the r and s representation because they are absolutely equivalent we know that r and s are expressed as a and b so it's easier for me to write it in this way as I did before because I know that r and s is still expressed as a b and c it's some numbers doesn't really matter a b and c numbers so s and r are also numbers and a so I will use this as a as a basis as a foundation for for this analysis graphical analysis let's draw this particular function on the coordinates x so step number one plane parabola so you know that this is an even function even which means it's symmetrical if this is minus one this is one this is one it goes like this this is y is equal to x square great what's next step next step is y is equal to x minus s square what does it mean well if you remember what happens with the graph of this function relative to this function well the whole graph is shifted towards the right so if this is s then the whole graph will look like this not very symmetrical let me try again maybe that's better so it's also symmetrical relative to this axis rather than this well why is that why is that well obviously if if function is x minus s square then it takes exactly the same value for x equal s as the original function takes with x equal to zero whatever this function takes in s plus one let's say is the same as the original one function in x is equal to one that's that everything is just shifted to the right by s okay that's great that's first transformation now the second transformation is multiplication by a okay what happens with the graph if we multiply it by a well if a is let's say two then we're stretching vertically um two times so zero remains zero but whatever was one for instance will become two whatever was four will become eight etc so graph will be steeper so it will be like this but that's only if a is positive if a is negative then everything is more or less the same but the horns of this parabola will go down that's the only difference but in any case it's vertically stretching eight times and again if a is positive the horns will remain going upwards and if a is negative the horn will the horns will reverse the direction so just for simplicity let's assume that a is positive and so this will be our graph okay now minus r so that's another transformation what happens with the function if we subtract r from it well obviously the graph will go down by r in this case and again let's assume r is positive then down means really down because if r is negative subtracting r means actually shifting it up so in this particular case again for simplicity let's assume that r is positive then this is minus r so the whole parabola will go this way okay now let me draw everything but the last piece because now it looks really messy so the last piece is zero x y this is s this is minus r this is the focal point of our graph and then it goes upwards this is our parabola now what are the roots of this equation equation of eight times x minus s square minus r is equal to zero well obviously that's where y is equal to zero this is y axis this is y is equal to zero so wherever this parabola crosses the the x axis that's two solutions well again that's a very simple case that's in case r is greater than zero and s is greater than zero now what happens if our r in this particular case is negative well if r is negative then the parabola will go instead of down will go up because minus and then the negative will be positive number so the whole graph will shift up now if the graph shifts up instead of down there will be no solutions and this is exactly equivalent to the case when we were saying that if you remember b square minus four ac will be greater or equal to zero r and s being expressed in terms of ab and c result in exactly the same type of restriction so this restriction for ab and c and these restrictions for r and s are really equivalent because if you will basically do the arithmetic back to whatever i was expressing r and s in terms of ab and c you will see that these two will probably be equivalent to something like this this is not the only kind of restrictions which we really have to make but the most important one is actually r restriction for r because s is just shift horizontally back and forth but r greater than zero in this particular case this is really supposed to be equivalent to this and if you remember what r actually was then you will see that that's exactly the case so having or not having solutions in the graphical form is quite obvious it's basically this parabola having these two quarts whether the quarts cross the x axis or they don't now one interesting point remember what if b square minus four ac is equal to zero if you remember this is under the square root in the formula of the solutions of this particular equation and from the formula it's actually one solution instead of two because it's plus square root and minus square root square root is equal to zero from this expression so basically it's the same solution now considering what this is actually representing it represents r so if r is equal to zero the parabola doesn't shift from the original position so instead of going upward down if r is equal to zero it stays it stays put vertically and it's still this and it has only one touching point so that's the only point where we have only one solution so as soon as r is positive or which is the same thing b square minus four ac greater than zero parabola goes down and there are two solutions in case r is negative parabola goes up and there is no solution and finally if r is equal to zero there is only one solution so this r which is related to this formula is exactly the value of this r determines whether parabola will or will not have any solutions or if it does that it's one or two solutions by the way considering if solutions are either are two solutions what happens with parabola being a little bit lower like this and then moves up which means r is diminishing it's in its value the parabola will move up and these two roots will start approaching each other so there are still two all the time up until the point when r is exactly equal to zero that's why in this particular case sometimes this solution is called a double point basically yes it's still one point but it represents a limit case for all those parabolas where there are really two points of crossing the x-axis so sometimes even the solution is only single in case this is equal to zero we are saying that this is a double solution just to basically represents that this is a limiting case well that concludes our lecture about quadratic equations there will be some exercises and maybe we'll spend some time in the future for complex solutions to quadratic equations well that's it for today thank you very much