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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.

## Want to join the conversation?

• At Sal factors out a 4 to simplify the equation, which I understand. I also understand that, in the end, this equation doesn't have any real roots because we can't take the square of a negative. However, wouldn't the final answer be -45 = 4(x+5)^2 instead of -45 = (x+5)^2. The 4 that he factored out seems to have disappeared into the ether. Am I missing something?
(7 votes)
• John,
Rather that "factor out the 4", Sal divided both sides of the equation by 4. On the left was a zero, and 0/4 = 0, so yes, the 4 did look like it evaporated into thin air.
But if he had just factored the 4, the answer would be
-180 = 4(x+5)². And we could still then divide both sides by 4 and the answer would be -45 = (x+5)²

0=4x² + 40x +280 Let's just factor out the 4
0=4(x² + 10x + 70) And and and subtract the 25 inside the parenthesis.
0=4(x² + 10x + 25 - 25 + 70) Now convert the perfect square
0=4((x+5)² -25+70) subtract the 25 from 70
0=4((x+5)² + 45) Distribute the 4
0=4(x+5)² + 4*45 and multiply 4*45
0=4(x+5)² + 180 Now subtract 180 from each side
-180 = 4(x+5)² And now you could divide each side by 4
-45 = (x+5)²

So, as you see the 4 didn't just disappear. It is that when one side was 0, when both sides were divided by 4 it just looked like it was factored out and disappeared.

I hope that makes it click for you.
(17 votes)
• Does the square root always have to equal zero
(8 votes)
• When finding the roots, solutions, or zeros (all the same thing) to a quadratic function, yes, you set them equal to zero, and solve for the independent variable (x).
(13 votes)
• What would the complex roots be?
(6 votes)
• Hello gatogreensleeves,
the complex roots would be x = -5 +j*(squareroot(45)) and x = -5 -j*(squareroot(45)).
Regards!
(6 votes)
• why can't you use x = -b + or - sqrt b squared - 4ac/2a for this?
(4 votes)
• You certainly can use that formula (which is called the quadratic formula). However, Sal is trying to explain HOW to find the solutions of the equation so that people will understand how equations are manipulated. Simply plugging numbers into a formula certainly works, but it doesn't help with understanding and it makes quite a boring video.
(11 votes)
• At , Sal said we can add a billion and subtract a billion without changing the equation. I agree completely with that.
But can we add infinite and subtract infinite without changing the equation?
(1 vote)
• No, because you can't actually add or subtract infinity. It is not a number, it is an abstract concept.

Infinity minus infinity is not zero. Similarly, infinity divided by infinity is not one. Therefore, you cannot apply the same rules of inverse operations to infinity as you would to an actual number.
(9 votes)
• at what is a complex number?
(4 votes)
• Complex numbers are a new class of numbers that support roots of negative numbers.
They are written in the form a + bi where i is sqroot(-1). Note that complex numbers include the real numbers; you could think of a real number as being a complex number where b=0.
In the UK they are introduced at A-level, and used extensively in engineering degree calculations. At GCSE it is sufficient to state that there are no real roots. If you were to sketch the graph, it would never intersect the x-axis (i.e. there is no real value of x such that y=0).
(3 votes)
• When talking about the complex roots would you need a 3 dimensional graph to plot the roots? With x, y and i axes? Or is this not possible? If not why not?
(0 votes)
• Generally complex numbers are graphed on a slightly modified coordinate plane with real numbers represented by what we usually think of as the x-axis and imaginary numbers represented by what is usually thought of as the y-axis. For example, the number 3 + 4i on the complex plane would be in the same place as the point (3, 4) on the x-y plane.

That said, I don't see why what you propose wouldn't be a reasonable representation. It'd certainly be interesting to see how the graphs look with both real and complex roots showing!
(10 votes)
• My professor said to find the vertex of the parabola y=x^2-4x+6. When you complete the square you end up getting an answer that has no solution as Sal showed above. But then why is the vertex (2,2)?

Thank You!
(2 votes)
• The vertex is different than the solution. The vertex is the maximum/minimum point of the line, and the solution(s) is where the line crosses the x-axis. The vertex can be found using this equation: x=(-b)/2a. This will get you the x value of the vertex. Plug that into your original equation to find y, and you have the coordinates of the vertex. :)
(3 votes)
• Do we follow BODMAS in reverse for algebraic manipulation? I don't see why he couldn't have taken the square root of both sides before moving 45 to the left. He'd have had: -sqrt(45) = (x + 5).
(2 votes)
• When you perform algebraic manipulation like this, you have to remember to perform the same operation to the entire side on both sides. That means taking the square root of both sides at that point, you'd have to take the square root of (x+5)^2 + 45 as a whole, not just the (x+5)^2 part. Then with the 45 "stuck" under the square root, you wouldn't be able to just subtract 45 from each side easily.

You don't have to follow any particular order for which operations you perform during manipulation. Generally you need to look at what you have and see which operation will help you get closer to your goal, and remember, you have to perform the operation on the entire side as a whole for both sides.
(2 votes)
• I see how it fits, but why did we take 1/2 of b and square it?
(2 votes)
• Great question
Sal is trying to complete the square and in order to do that he wants the right side to look something like this:
(ax + c)^2 where 'a' and 'c' can be any number
If we work backwards it might make more sense (going from (a + b)^2 to what Sal starts with).
(ax + c)^2 ----> (ax)^2 + 2acx + c^2
We have 'a' (a = 1), and we know that b = 2ac, so to complete the square we just have to find c^2
b = 2ac ----(a = 1)----> b = 2c ----> b/2 = c ----> (b/2)^2 = c^2
I hope that helps!
(2 votes)

## 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.