If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

Quadratic systems: a line and a parabola

A system of equations that contains one linear equation and one quadratic equations can be solved both graphically and algebraically. Each method has its pros and cons. See an example using both methods.

Want to join the conversation?

Video transcript

- [Instructor] We're told the parabola given by y is equal to three x squared minus six x plus one and the line given by y minus x plus one equals zero are graphed. So we can see the parabola here in red, and we can see the line here in blue. And the first thing they ask us is one intersection point is clearly identifiable from the graph. What is it? And they want us to put it in here. This is actually a screenshot from the exercise on Khan Academy, but I'm just going to write on it. If you were doing it on Khan Academy, you would type it in. But pause this video and see if you can answer this first part. All right, so one intersection point is clearly identifiable from the graph. I see two intersection points. I see that one, and I see that one there. This second one seems clearly identifiable because, when I look at the grid, it looks clearly to be at a value of x equals two and y equals one. It seems to be the point two comma one. So it's two comma one there. And what's interesting about these intersection points is, because it's a point that sits on the graph of both of these curves, that means that it satisfies both of these equations, that it's a solution to both of these equations. So the other one is find the other intersection point. Your answer must be exact. So they want us to figure out this intersection point right over here. Well to do that, we're going to have to try to solve this system of equations, and this is interesting because this is a system of equations where one of the equations is not linear, where it is a quadratic. So let's see how we could go about doing that. So let me write down the equations. I have y is equal to three x squared minus six x plus one. And our next equation right over here, y minus x plus one is equal to zero. Well one way to tackle, and this is one way to tackle any system of equations, is through substitution. So if I can rewrite this linear equation as in terms of y, if I can solve for y, then I can substitute what y equals back into my first equation, into my quadratic one, and then hopefully I can solve for x. So let's solve for y here. And actually, let me color code it, because this one is in red, and this one is the line in that blue color. So let's just solve for y. The easiest way to solve for y is to add x to both sides and subtract one from both sides. That was hard to see. So and subtract one from both sides. And so we are going to get y, and then all the rest of the stuff cancels out, is equal to x minus one. And so now we can substitute x minus one back in for y, and so we get x minus one is equal to three x squared minus six x plus one. Now we wanna get a zero one side of this equation, so let's subtract x. I'll do this in a neutral color now. Let's subtract x from both sides. And let's add one to both sides. And then whatta we get? On the left-hand side we just get zero. And on the right-hand side we get three x squared minus seven x plus two. So this is equal to zero. Now we could try to factor this. Let's see, is there an obvious way to factor it? Can I think of two numbers, a times b, that's equal to the product of three and two? Three times two, and if this looks unfamiliar, you can review factoring by grouping. And can I think of those same two, a plus b, where it's going to be equal to negative seven? And actually, negative six and negative one work. So what can do is I can rewrite this whole thing as zero is equal to three x squared, and then instead of negative seven x, I can write negative six x and minus x. And then I have my plus two. I'm just factoring by grouping, for those of you are not familiar with this technique. You could also use the quadratic formula. So then zero is equal to, so if I group these first two I can factor out a three x. So I'm going to get three x times x minus two, and in these second two I can factor out. In these second two I can factor out a negative one, so I have negative one times x minus two. And then I can factor out a negative two. I'll scroll down a little bit so I have some space. So I have zero is equal to, if I factor out an x minus two, I'm going to get x minus two times three x minus one. And so a solution would be a situation where either of these is equal to zero, or, I'll scroll down a little bit more. So x minus two could be equal to zero, or three x minus one is equal to zero. The point at which x minus two equals zero is when x is equal to two. And for three x minus one equals zero, add one to both sides. You get three x is equal to one, or x is equal to 1/3. So we figured out the, we already saw the solution where x equals two. That's this point right over here. We already typed that in. But now we figured out the x value of the other solution. So this is x is equal to 1/3 right over here. So our x value is 1/3. But we still have to figure out the y value. Well the y value's going to be the corresponding y we get for that x in either equation. And I like to focus on the simpler of the two equations. So we could figure out what is x when, or what is y when x is equal to 1/3 using this equation? We could have used the original one, but this is even simpler. It's already solved for y. So y is equal to 1/3 minus one. I'm just substituting that 1/3 back into this. And so you get y is equal to negative 2/3. And it looks like that as well. y is equal to negative 2/3 right over there. So is the point 1/3 comma negative 2/3. And we're done.