Negative 2- same idea there- and then you square it. Half of that coefficient? Well, half of that coefficient You literally just look at thisĬoefficient right here, and you say, OK, well what's And this might look allĬomplicated to you right now, but I'm showing you Negative 2, what is a squared? Well, then a squared is going So this number, a is going toīe half of negative 4, a has to be negative 2, right? Because 2 times a is going Right here, that has to be- sorry, x squared minus 2ax. Is going to be x squared minus 2a plus a squared. Is just yet, but we know a couple of things. Something to be equal to x minus a squared. Think it'll become clear with a few examples. Then if I have two times my number I get negative 4. That if I have my number squared I get that number, and Left-hand side to be a perfect square, there has to be So how can we do that? Well, in order for this Perfect square, engineering it, adding and subtracting fromīoth sides so it becomes a perfect square. ![]() Is all about making the quadratic equation into a That these can be pretty straightforward to solve if Quadratic equation x squared minus 4x is equal to 5. Of what we did in the last video, where we solved Need to understand even what it's all about. Video after that I'll prove the quadratic formula usingĬompleting the square. ![]() That this will work for any quadratic equation, and it'sĪctually the basis for the quadratic formula. You a technique called completing the square. Sorry for the large number of words used to answer your question. Sal uses (x - a)^2 simply to tell this is the format of the factorised form of : If it is there then you have to divide the whole equation first by the coefficient and then halve the coefficient of x and put it into the brackets. But remember, you only halve the coefficient of x and put it into the brackets only if there is no number before x^2 (coefficient of x^2). Where a is half the coefficient of x(the number before the x), as long a there is no coefficient of x^2. You want x^2 - 4x + (something) to be equal to (x-a)^2. (x-a)^2 is basically the format of the answer you want receive. Notice that now I could factorise the left hand side into (x - 2)^2 I want to factorise the left side of the equal sign, so I have to find a value for (something) which would allow me to factorise the left-hand side of the equation. Situations could vary, but this is the basic idea behind the procedure. Finally, you add 1 to both sides, taking into account that 4 could be positive OR negative. Next, you want to take the square root from both sides so that x-1 is equal to the positive or negative square root of 16 (positive or negative 4). This will all give you the equation (x-1)^2=16. Because you´re taking this value away from the constant, you will add it to the other side of the equation (this might not make sense at first, but if the constant were on the variable side, you would be subtracting). In this case, you will add 1 because it perfectly factors out into (x-1)^2. Next, you want to add a value to the variable side so that when you factor that side, you will have a perfect square. Because there is a 3 in front of x^2, you will divide both sides by 3 to get x^2-2x=5. Next, you want to get rid of the coefficient before x^2 (a) because it won´t always be a perfect square. To do this, you will subtract 8 from both sides to get 3x^2-6x=15. To complete the square, first, you want to get the constant (c) on one side of the equation, and the variable(s) on the other side. ![]() You don´t need another video because I´m about to explain it to you! Say you have the equation 3x^2-6x+8=23.
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