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# Solving quadratics by completing the square: no solution

CCSS Math: HSA.REI.B.4, HSA.REI.B.4a, HSA.REI.B.4b, HSA.SSE.B.3, HSA.SSE.B.3b

## 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.