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## Integrated math 3

### Course: Integrated math 3>Unit 7

Lesson 5: Quadratic systems

# Quadratic system with no solutions

Sal solves a system of two quadratic equations algebraically and finds the system has no solutions. He then graphs the equations to show that this is true. Created by Sal Khan and Monterey Institute for Technology and Education.

## Want to join the conversation?

• Is there a video on how to put an equation in the vertex form? It's appears like easy manipulation however, is there something that could help me better understand how Sal played those equations so eloquently?
(21 votes)
• 18 / 0.12= 150. How do you get that?
(7 votes)
• an easy way to think about it is that you want to clear away the decimal. so by moving the decimal from .12 two places to the right (or multiply by 100). and because you moved it to the right for .12 you also want to do that to the 18 to make it 1800. 1800/12=150.
(15 votes)
• Would the two parabolas intersect when using an imaginary axis?
(8 votes)
• The imaginary axis is a completely different dimension than the Cartesian Plane. It is used to visualize an imaginary point, so technically you can't graph a parabola on an imaginary axis. The two parabolas from the video won't intersect on the real axis, which means they won't intersect at all.
I hope that helped! 🙂
(5 votes)
• @ Isn't it incorrect to say there is no solution at all? Is there not a solution that just involves i?
(6 votes)
• Yeah, but he wants to find all the real solutions. There are solutions that involve i, but they are not real solutions.
(8 votes)
• can someone please explain how 2(x-4)^2+3 turns into 2(x^2-8x+16)+3?
(7 votes)
• (x-4)^2 =(x-4)(x-4)
You need to foil or multiply both together.
(x-4)(x-4)
Whenever you see two numbers in parenthesis being added or subtracted, you need to multiply the terms together:
Ex:
(x+3)^2=(x+3)(x+3)
(4x-2)^2=(4x-2)(4x-2)

Foiling
(x-4)(x-4)
Multiply 1st terms (x*x)
x^2
Multiply 1st term to last term
(x)(-4)
-4x
Multiply middle terms
(-4)(x)
Multiply last terms
(-4)(-4)=16
x^2-4x-4x+16
x^2-8x+16
You can also just distribute the terms
(x-4)(x-4)
Distribute either one. I will distribute the x-4 to the other x-4

x(x-4) - 4(x-4)
x^2-4x - 4x+16
Combine like terms
x^2-4x-4x+16
x^2-8x+16
(5 votes)
• so should i have studied quadratic equations before i got to this? it looks like that's a later topic in the algebra play list. and where do i find the vertex form/formula?

for instance at sal says "this is in vertex form" referring to y = 2 (x-4)^2 + 3 "where x = 4 and y = 3"... and then he plots that point. i don't knw what that means or where i can look for it in the playlist.

at i think this might be completing the square?
(4 votes)
• Yes, you need to know quadratics before doing these videos -- they seem to have been put too early on this list. It is possible to do some simple non-linear equations before mastering quadratics, but I would not recommend it. Try going to the Functions videos and then coming back to these videos once you're completely done with quadratics.
(4 votes)
• What if in one or both of the equations y and x are squared?
(3 votes)
• You'd have to simplify until the equation is either linear or quadratic.
Think of it like this
ax^2+bx+c = a ( Substitute ^2 ) ^2 + b ( Substitute ^2 ) + c

If an equation is just y^2 = ax^2+bx+c
convert it to y = sqrt(a) * x + sqrt (b) sqrt ( x ) + sqrt(c) and then insert this into the other equation and solve.
(2 votes)
• How would a quadratic look if it had 1 solution?
How can you manipulated?
(2 votes)
• You get one solution when the discriminant b²-4ac=0. The graph of the parabola would have the vertex on the x-axis, so that it only touches at one point.
(4 votes)
• Hello, I have a question about Quadratic system with a wrong solution.
If a quadratic system consists:
Equation 1: ax^2 = y
Equation 2: bx = y
for any real number a & b.
It will yield:
ax^2 = bx
ax^2 - bx = 0
x(ax-b) = 0
Hence, x = 0 & y = 0 Or x = b/a & y = b^2/a
Although we can simply find x = 0 & y = 0 is the true answer (by plotting a graph), what is the reason/meaning behind of the equation ax-b = 0 (why does it cause)? Is this mathematic fault in the quadratic system unevitable? Sorry for bothering and many thanks.
(2 votes)
• What leads you to believe that x = b/a & y = b^2/a is not a solution? It is a valid solution. If you substitute it back into both equations, the ordered pair works in both equations.
(3 votes)
• At , why did he add the "plus 1 and minus 1" there? Won't it be cancelled out and therefore he would be adding nothing to the equation? Is there something I'm missing or not?

Please help me with this. This doesn't make sense.
(3 votes)
• Yes, your observation is correct.
To understand what Sal is doing, we have to see where Sal wants to go. We have a function like y=ax^2+bx+c, where a=1.

One way (shown in the video) to simplify this is to isolate the terms with an 'x'. A formula to look into is (x+(1/2)b)^2= x^2+bx+b^2
To reconstruct x^2-2x+2 to fit this formula (x+(1/2)b)^2, we have x^2-2x+c (ignore c, for now). rearranging this into (x-1)^2, which is equivalent to x^2-2x+1+c; here you can see we have 1 'too much'. Therefore we have to subtract 1 from 'c'. (see video). This indeed looks weird as we have x^2-2x+1+2-1.
(1 vote)

## Video transcript

Solve the system of equations using any method. We have y is equal to 2 times the quantity x minus 4 squared plus 3. We also have y is equal to negative x squared plus 2 x minus 2. The solution-- it might be one, it might be none, or it might be two solutions-- to this system occurs for the x values that generate the same y values. There's the same x and y that satisfy both of these equations. In order to find the x values, they need to equal the same y values, so this y has to be that y value. So the solution is going to occur when this guy right here-- negative x squared plus 2x minus 2 is equal to that guy up there, or equal to 2 times x minus 4 squared plus 3. Now let's just try to solve for x. The left hand side-- we're going to have to multiply this out, so let's do that first. It's negative x squared plus 2x minus 2 is equal to. And on the right hand side, 2 times x minus 4 squared is x squared minus 8x plus 16 plus 3. This is going to be equal to 2x squared-- I'm just distributing the 2-- minus 16x plus 32 plus 3, which is equal to 2x squared minus 16x plus 35. That's, of course, going to be equal to this thing on the left hand side, negative x squared plus 2x minus 2. Let's just get rid of this whole thing from the left hand side all at once by adding x squared to both sides. We can all do it in one step. We're going to add x squared to both sides. Let's subtract 2x from both sides, and let's add 2 to both sides. On the left hand side, those cancel out, those cancel out, those cancel out. You're left with 0 is equal to 2x squared plus x squared is 3x squared. Negative 16x minus 2x is negative 18x, and then 35 plus 2 is 37. So we just have a plain vanilla quadratic equation right here. We might as well apply the quadratic formula here to try to solve it. Our solutions are going to be x is equal to negative b. Well, b is negative 18, so negative b is positive 18. It's 18 plus or minus the square root of 18 squared minus 4 times 3 times c-- times 37. All of that is over 2 times a-- 2 times 3, which is 6. Let's think about what this is going to be. Over here, we have 18 plus or minus the square root of-- let's just use a calculator. I could multiply it out but I think-- we have 18 squared minus 4 times 3 times 37, which is negative 120. It's 18 plus or minus the square root of negative 120. You might have even been able to figure out that this is negative. 4 times 3 is 12. 12 times 37 is going to be a bigger number than 18. Although it's not 100% obvious, but you might be able to just get the intuition there. We definitely end up with a negative number under the radical here. Now, if we're dealing with real numbers, there is no square root of negative 120. So there is no solution to this quadratic equation. There is no solution. If we wanted to, we could have just looked at the discriminant. The discriminant is this part-- b squared minus 4ac. We see the discriminant is negative, there's no solution, which means that these two guys-- these two equations-- never intersect. There is no solution to the system. There are no x values that when you put into both of these equations give you the exact same y value. Let's think a little bit about why that happened. This one is already in kind of our y-intercept form. It's an upward opening parabola, so it looks something like this. I'll do my best to draw it-- just a quick and dirty version of it. Let me draw my axes in a neutral color. Let's say that this right here is my y-axis, that right there is my x-axis. x and y. This vertex-- it's in the vertex form-- occurs when x is equal to 4 and y is equal to 3. So x is equal to 4 and y is equal to 3. It's an upward opening parabola. We have a positive coefficient out here. So this will look something like this. I don't know the exact thing, but that's close enough. Now, what will this thing look like? It's a downward opening parabola and we can actually put this in vertex form. Let me put the second equation in vertex form, just so we have it. So we have a good sense. So, y is equal to-- we could factor in a negative 1-- negative x squared minus 2x plus 2. Actually, let me put the plus 2 further out-- plus 2, all the way up out there. Then we could say, half of negative 2 is negative 1. You square it, so you have a plus 1 and then a minus 1 there. This part right over here, we can rewrite as x minus 1 squared, so it becomes negative x minus 1 squared. Let me just do it one step at a time. I don't want to skip steps. Negative x minus 1 squared minus 1 plus 2. So that's plus 1 out here. Or if we want to distribute the negative, we get y is equal to negative x minus 1 squared minus 1. Here the vertex occurs at x is equal to 1, y is equal to negative 1. The vertex is there, and this is a downward opening parabola. We have a negative coefficient out here on the second degree term, so it's going to look something like this. So as you see, they don't intersect. This vertex is above it and it opens upward. This is its minimum point. And it's above this guy's maximum point. So they will never intersect, so there is no solution to this system of equations.