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Use completing the square to find the roots of the

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quadratic equation right here.

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And when anyone talks about roots, this just means find

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the x's where y is equal to 0.

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That's what a root is.

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A root is an x value that will make this quadratic function

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equal 0, that will make y equal 0.

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So to find the x's, let's just make y equal 0 and

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then solve for x.

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So we get 0 is equal to 4x squared plus 40x, plus 280.

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Now, the first step that we might want to do, just because

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it looks like all three of these terms are divisible by

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4, is just divide both sides of this equation by 4.

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That'll make our math a little bit simpler.

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So let's just divide everything by 4 here.

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If we just divide everything by 4, we get 0 is equal to x

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squared plus 10x, plus-- 280 divided by 4 is 70-- plus 70.

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Now, they say use completing the square, and actually, let

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me write that 70 a little bit further out, and you'll see

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why I did that in a second.

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So let me just write a plus 70 over here, just to have kind

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of an awkward space here.

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And you'll see what I'm about to do with this space, that

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has everything to do with completing the square.

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So they say use completing the square, which means, turn

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this, if you can, into a perfect square.

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Turn at least part of this expression into a perfect

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square, and then we can use that to actually solve for x.

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So how do we turn this into a perfect square?

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Well, we have a 10x here.

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And we know that we can turn this into a perfect square

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trinomial if we take 1/2 of the 10, which is 5, and then

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we square that.

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So 1/2 of 10 is 5, you square it, you add a 25.

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Now, you can't just willy-nilly add a 25 to one

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side of the equation without doing something to the other,

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or without just subtracting the 25 right here.

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Right?

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Think about it, I have not changed the equation.

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I've added 25 and I've subtracted 25.

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So I've added nothing to the right-hand side.

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I could add a billion and subtract a billion and not

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change the equation.

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So I have not changed the equation at all right here.

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But what I have done is I've made it possible to express

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these three terms as a perfect square.

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That right there, 2 times 5 is 10.

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5 squared is 25.

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So that is x plus 5 squared.

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And if you don't believe me, multiply it out.

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You're going to have an x squared plus 5x, plus 5x,

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which will give you 10x, plus 5 squared, which is 25.

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So those first three terms become that, and then the

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second two terms, right there, you just add them.

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Let's see, negative 25 plus 70.

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Let's see, negative 20 plus 70 would be positive 50, and then

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you have another 5, so it's plus 45.

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So we've just algebraically manipulated this equation.

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And we get 0 is equal to x plus 5 squared, plus 45.

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Now, we could've, from the beginning if we wanted, we

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could've tried to factor it.

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But what we're going to do here, this will always work.

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Even if you have crazy decimal numbers here, you can solve

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for x using the method we're doing here,

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completing the square.

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So to solve for x, let's just subtract 45 from both sides of

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this equation.

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And so the left-hand side of this equation becomes negative

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45, and the right-hand side will be just

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the x plus 5 squared.

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These guys, right here, cancel out.

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Now, normally if I look at something like this I'll say,

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OK, let's just take the square root of both

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sides of this equation.

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And so you might be tempted to take the square root of both

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sides of this equation, but immediately when you do that,

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you'll notice something strange.

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We're trying to take the square root

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of a negative number.

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And if we're dealing with real numbers, which is everything

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we've dealt with so far, you can't take a square root of a

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negative number.

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There is no real number that if you square it will give you

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a negative number.

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So it's not possible-- I don't care what you make x-- it is

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not possible to add x to 5 and square it, and

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get a negative number.

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So there is no x that can satisfy-- if we assume that is

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x is a real number-- that can satisfy this equation.

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Because I don't care what x you put here, what real x you

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put here, you add 5 to it, you square it, there's no way

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you're going to get a negative number.

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So there's no x that can satisfy this equation, so we

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could say there are no-- and I'm using the word real

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because in Algebra 2 you'll learn that there are things

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called complex numbers, but don't worry about that right

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now-- but there no real roots to the quadratic equation.

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And we're done.

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And actually, if you had tried to factor it, you would have

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found it very difficult, because this is not a

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factorable expression right here, and you know it because

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there's no real roots.

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