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# Solving quadratics by completing the square: no solution

Sal solves the equation 4x^2+40x+280=0 by completing the square, only to find there's no solution for this equation. Created by Sal Khan and Monterey Institute for Technology and Education.
Video transcript
Use completing the square to find the roots of the quadratic equation right here. And when anyone talks about roots, this just means find the x's where y is equal to 0. That's what a root is. A root is an x value that will make this quadratic function equal 0, that will make y equal 0. So to find the x's, let's just make y equal 0 and then solve for x. So we get 0 is equal to 4x squared plus 40x, plus 280. Now, the first step that we might want to do, just because it looks like all three of these terms are divisible by 4, is just divide both sides of this equation by 4. That'll make our math a little bit simpler. So let's just divide everything by 4 here. If we just divide everything by 4, we get 0 is equal to x squared plus 10x, plus-- 280 divided by 4 is 70-- plus 70. Now, they say use completing the square, and actually, let me write that 70 a little bit further out, and you'll see why I did that in a second. So let me just write a plus 70 over here, just to have kind of an awkward space here. And you'll see what I'm about to do with this space, that has everything to do with completing the square. So they say use completing the square, which means, turn this, if you can, into a perfect square. Turn at least part of this expression into a perfect square, and then we can use that to actually solve for x. So how do we turn this into a perfect square? Well, we have a 10x here. And we know that we can turn this into a perfect square trinomial if we take 1/2 of the 10, which is 5, and then we square that. So 1/2 of 10 is 5, you square it, you add a 25. Now, you can't just willy-nilly add a 25 to one side of the equation without doing something to the other, or without just subtracting the 25 right here. Right? Think about it, I have not changed the equation. I've added 25 and I've subtracted 25. So I've added nothing to the right-hand side. I could add a billion and subtract a billion and not change the equation. So I have not changed the equation at all right here. But what I have done is I've made it possible to express these three terms as a perfect square. That right there, 2 times 5 is 10. 5 squared is 25. So that is x plus 5 squared. And if you don't believe me, multiply it out. You're going to have an x squared plus 5x, plus 5x, which will give you 10x, plus 5 squared, which is 25. So those first three terms become that, and then the second two terms, right there, you just add them. Let's see, negative 25 plus 70. Let's see, negative 20 plus 70 would be positive 50, and then you have another 5, so it's plus 45. So we've just algebraically manipulated this equation. And we get 0 is equal to x plus 5 squared, plus 45. Now, we could've, from the beginning if we wanted, we could've tried to factor it. But what we're going to do here, this will always work. Even if you have crazy decimal numbers here, you can solve for x using the method we're doing here, completing the square. So to solve for x, let's just subtract 45 from both sides of this equation. And so the left-hand side of this equation becomes negative 45, and the right-hand side will be just the x plus 5 squared. These guys, right here, cancel out. Now, normally if I look at something like this I'll say, OK, let's just take the square root of both sides of this equation. And so you might be tempted to take the square root of both sides of this equation, but immediately when you do that, you'll notice something strange. We're trying to take the square root of a negative number. And if we're dealing with real numbers, which is everything we've dealt with so far, you can't take a square root of a negative number. There is no real number that if you square it will give you a negative number. So it's not possible-- I don't care what you make x-- it is not possible to add x to 5 and square it, and get a negative number. So there is no x that can satisfy-- if we assume that is x is a real number-- that can satisfy this equation. Because I don't care what x you put here, what real x you put here, you add 5 to it, you square it, there's no way you're going to get a negative number. So there's no x that can satisfy this equation, so we could say there are no-- and I'm using the word real because in Algebra 2 you'll learn that there are things called complex numbers, but don't worry about that right now-- but there no real roots to the quadratic equation. And we're done. And actually, if you had tried to factor it, you would have found it very difficult, because this is not a factorable expression right here, and you know it because there's no real roots.