Algebra

Solving a quadratic by factoring | Quadratic equations | Algebra I | Khan Academy

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Welcome to display the solution of the quadratic equation by extracting the factor Let’s start with solving a few issues Let’s assume I have the conjugation f (x) = x ^ 2 + 6x + 8 Now if you want to graphically represent (f) x, represent it It will look like this I don’t know exactly what it will look like, but it is Will be parabola and will cross the x axis In two points, here and here What we will try to do is define These two points First, when a coupling crosses with X axis, then f (x) = 0 Because this f of the x axis, is similar to the y axis So here f (x) = 0 To solve this equation, we will set f (x) = 0 And we get x ^ 2 + 6x + 8 = 0 Now perhaps this solution seems simple But x ^ 2 makes things messy and you can Try to solve it yourself What we will do is analyze it to its factors And we will say x ^ 2 + 6x + 8 You can write as (x + something) x (x + something) And it’s still equal to that, except it’s 0 Now in these brackets, I will clarify Systematic method or you can say mechanical method of doing that I will give you another lecture on why this works Perhaps you want to hit the answers We will start and multiply the phrases and We see the reason for its success And what we’re going to use here is, look at The coefficient of this x expression is 6 And we say what are the two numbers that total 6 And when we hit them We will get 8 Well, let’s think about the 8 factors The factors for 8 are 1, 2, 4, and 8 Well, 1 x 8 = 8, but 1 + 8 = 9, so this doesn’t work 2 x 4 = 8, and 2 + 4 = 6 This worked We can say x + 2 and x + 4 equals 0 Now, if two phrases or two numbers are multiplied together, they will be summed up 0, it means that one or both of these numbers are present It should equal 0 Now we can say x + 2 = 0, and x + 4 = 0 Well, this is a very simple equation We subtract 2 from both sides and get x = -2 Here we get x = -4 And if we replace any of these in The original equation, we will see that this works -2 – So let’s try with -2 and I’ll leave -4 To you – So -2 ^ 2 + 6 x -2 + 8 -2 ^ 2 = 4, -12 – i 6 x -2– + 8 And certainly this is equal to 0 And if you do the same thing with -4 You will also see that this works And you might say now, that’s amazing, it’s fun This equation has two solutions Well, if you think about it, it would make sense because graphic representation For f (x) intersects the x-axis in two different locations Let us solve another issue Let’s assume I have f (x) = 2x ^ 2 + 20x + 50 And if you want to find the location of its intersection with the x axis Just put f (x) = 0, and it will reflect The left and right side of the equation And get 2x ^ 2 + 20x + 50 = 0 Now, the slight difference between this equation and that is The coefficient of x ^ 2 here is equal to 2 instead From 1, I prefer 1 So let’s divide the whole equation, from the left side And right over 2 So I get x ^ 2 + 10x + 25 = 0 All I did was hit 1/2 x – this same Divide by 2 – x 1/2 Of course, 0 x 1/2 = 0 Now we are ready to do as we did before, and Maybe you want to stop the show and try for yourself We will say (x + something) x (x +) Something) = 0 and these two unknowns Their sum should be 10, and when we multiply them They must have a score of 25 Let’s consider the factors of 25 We have 1, 5, and 25 Well, 1 x 25 = 25 1 + 25 = 26, not 10 5 x 5 = 25, and 5 + 5 = 10, so this worked out It turns out that both of these numbers are 5 We get x + 5 = 0 or x + 5 = 0 We will write it only once So we get x = -5 How do you think of this graphically? I told you that these equations can cross With x axis in two places, but this one has only one solution Well, this solution will look like If x = -5, then we will get parabola It touches the axis from here and then back Instead of intersection in two places, you will Intersect here only at x = -5 And now as an example proves to you that I am not Wrong notify you, let’s multiply (x + 5) x (x + 5) In order to show you that it is equal to what it must be equal to So we said that this equals x (x + 5) + 5 (x + 5) x ^ 2 + 5x + 5x + 25 This equals x ^ 2 + 10x + 25 So it is equal to what we said it should be equal I will design another lesson again, where I will explain This is more Let us solve another issue In this example, I will conclude the presentation Let’s solve x ^ 2 – x – 30 = 0 Again, we want two numbers when we add them together – the result – what is The modulus here, it’s -1 So we can say that these two numbers are a + b = -1, and a x b = -30 Well, let’s think about all of the 30 factors 1, 2, 3, 5, 6, 15, 30 Well, something interesting is happening right now Since a x b = -30, one of these numbers It must be negative They cannot both be negative, because if they are both Salban, therefore this would be positive 30 a x b = -30 In fact we will say, two of these factors The difference between them should be -1 Well, if we all look at them, all these numbers Clearly when you mate them, the product of each pair will be 30 But the only two whose difference equals 1 are 5 and 6 And since it’s -1, it’ll be– I know I will be very fast with this and I will do a solution Other issues – this will be (x – 6) x (x + 5) = 0 How do I think about this? -6 x 5 = -30 -6 + 5 = -1 So this worked The more you practice this kind of issue – know They look annoying– you’ll You see more So we get x = 6 or x = -5 I think you are at this point able to solve Quadratic equations by analyzing them to their factors and I will solve other problems Soon until you gain experience on these issues enjoy your time

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