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### Course: Algebra 2 > Unit 10

Lesson 5: Quadratic systems# 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?

- At5:20, why does solving for x using 0 gives us the intersecting points of the two equations? I would expect that solving for x when the equation equals zero, would give us x intercepts.(8 votes)
- I think i might understand why:

We are not finding where y = 0, which would give the x intercepts. We are solving a system of two equations by finding the y value of one equation (i.e y = x-3) and inputting it into the second equation. Then rearranging this new equation so that it can be solved using the zero product property.

Any elaboration/correction greatly appreciated.(9 votes)

- At3:48, where does he get 6 and -1 from? Does he get it from earlier in the video and I missed it or what?(3 votes)
- Please check out factoring by groups on this website called - Factoring quadratics: leading coefficient ≠ 1.

the short answer is; to find a common factor from the function ax^2+bx+c

- find two numbers d & e

that multiply equal a times c; -- d*e=a*c

that add up to b; -- d+e=b

In the video that are -6 & -1 as they multiply to get 3*2 and add up to -7(5 votes)

- Is it ok to skip the factoring and use the quadratic formula? Or is it important to learn this way?(2 votes)
- Both methods are valid, but learning how to factor that way is very valuable when time is limited, so it might be worthwhile to practice.(4 votes)

- At2:30, why does he solve for y? Wouldn't it be easier to substitute y in the linear equation for the quadratic expression?(2 votes)
- Yes that is also another fast way to do it. There are many ways to approach a problem and still get the same solution — it's ultimately up for preference.(3 votes)

- @4:06we are told that as an alternative to factoring by grouping we could. How would you do that? I attempted it and did not get either answer that he did by grouping method.(1 vote)
- The quadratic formula says:

𝑎𝑥² + 𝑏𝑥 + 𝑐 = 0 ⇒ 𝑥 = (−𝑏 ± √(𝑏² − 4𝑎𝑐))∕(2𝑎)

In our case 𝑎 = 3, 𝑏 = −7, and 𝑐 = 2,

so 𝑥 = (−(−7) ± √((−7)² −4⋅3⋅2))∕(2⋅3)

= (7 ± √(49 − 24))∕6

= (7 ± √25)∕6 = (7± 5)∕6

⇒

𝑥 = (7 − 5)∕6 = 2∕6 = 1∕3

OR

𝑥 = (7 + 5)∕6 = 12∕6 = 2

So, the factorization should be (𝑥 − 1∕3)(𝑥 − 2).

But because 𝑎 = 3, the factorization is actually

3(𝑥 − 1∕3)(𝑥 − 2) = (3𝑥 − 1)(𝑥 − 2)(4 votes)

- y-x+1=0

+x +x

-1 -1

y = x-1(1 vote)

- At about4:30, how does the -1 magically appear? I don't see how Sal got it.(1 vote)
- Sal factors a -1 out of the expression. It is originally (-x + 1), and he rewrites it as -1(x - 1). You can imagine that he is 'un-distributing' a -1 to simplify the expression.(3 votes)

- i dont get it how do i do it(2 votes)
- Just do your best and eventually you will get it.(1 vote)

- 2:05

Can't you just substitute 3x^2-6x+1 for y in the linear equation?(1 vote)- Yes that would be quicker and more direct than what Sal did, you get to straight to 3x^2-6x+1-x+2=0, so 3x^2-7x+2=0.(1 vote)

- I can't undestand how solving for y in the linear equation can help me solve for y in the quadratic equation. Both equations have two intersection points, but most of them are different, why I can simply solve for Y?

I'm pretty confused on the "why I do it" even if technically I can do it.(1 vote)- ? What are you talking about? Solving for Y in the linear equation allows you to substitute it in the quadratic equation(1 vote)

## 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.