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## Algebra (all content)

### Course: Algebra (all content) > Unit 9

Lesson 11: Systems of quadratic equations# Quadratic systems: a line and a circle

Sal solves the system y=x+1 and x^2+y^2=25. Created by Sal Khan.

## Want to join the conversation?

- At0:55, Sal draws a circle, but from an equation like the one in the video, is there any way to figure out exactly how large the circle is?(14 votes)
- You can know it by substituting y by 0, so x will have two solutions on the x-axis, one is positive and the other is negative. The same would be if you substituted x by 0, y will have two solutions on the y-axis one is positive and the other is negative.(5 votes)

- 2:12How does he get the 2x?(7 votes)
- (x+1)² = (x+1)(x+1)

Use FOIL to get (x)(x) + 1x + 1x + (1)(1) = x² + 2x + 1(7 votes)

- At3:18of the video, how does Sal get that x= –4 or x= 3 from the "(x+4)(x-3)=0?"(4 votes)
- When we get down to the factored formula (x+4)(x-3)=0, then we use the
**zero property principle**that the others have described to finish the solution. The next step is to set each factor equal to zero by itself and solve.

(x+4) = 0

x + 4 -4 = 0 -4`x = -4 First solution for x`

(x-3) = 0

x-3 +3 = +3`x = 3 Second solution for x`

(10 votes)

- I can easily do those in my head, and so can many others. Why do we have to do all this work when It could be so much easier?(0 votes)
- With simple examples it may seem like we are hitting thumbtacks with a 30 pound sledgehammer rather than just pushing them in with our thumb. The point is to teach you, and get you used to, a method of solving equations, a method that works when you can't do the problems in your head because they are too complicated. You may be a fortunate person in that these processes are easy for you, but there are many others here for which this is hard - so they need this.(24 votes)

- Could we also use elimination like this?
`Turn y = x+1 into y^2 = x^2 + 2x + 1,`

Turn x^2 + y^2 = 25 into y^2 = 25 - x^2,

Subtract to get (x^2 + 2x + 1) - (25-x^2) = 0,

Simplify to the quadratic 2x^2 +2x - 24 = 0,

and continuing as shown in the video?(6 votes)- Yes, that works... you have the same quadratic to solve as Sal.(5 votes)

- why can one not simply sqrt x^2+(x+1)^2=25 and get x+x+1=5? I mean isn't the sqrt of x^2 =x and the sqrt of (x+1)^2 = x+1 and the sqrt of 25 = 5?(3 votes)
- √(a+b+c+...) is not equal to √(a)+√(b)+√(c)+...

So, √(x^2 + (x+1)^2) does not equal √(x^2) + √((x+1)^2)(5 votes)

- Hrm. I am a little disappointed to find out this only covers substitution of simpler algebraeic models. My question is: How do you solve the very most general models? Something like 3 = A(e^2B) + C, 7 = A(e^4B) + C and 9 = A(e^6B) + C using some numerical method. I don't think Khan Academy actually covers this. basically regression of nonlinear models using advanced numerical methods(4 votes)
- I can show you the methods for solving problems like that, if you need it. But they do get rather complicated.(3 votes)

- Could you please give the formula to turn "x^2 + y^2 = 25" into a circle with it's center at 0 and radius of 5? I know it's not directly related to the solution(s), but I would still like to know it. Thanks!(0 votes)
- You don't 'turn' that equation into a circle with radius 5 and center (0,0)... it already is that circle.

The general form for the equation of a circle is:

(x-h)^2 + (y-k)^2 = r^2 is the equation of a circle with center at (h,k) and radius r.

So, (x-4)^2 + (y+2)^2 = 49 has h=4, k=-2 and r=7, so it is the equation of a circle with center at (4, -2) and radius 7

For x^2 + y^2 = 25, think of it as (x-0)^2 + (y-0)^2 = 5^2, so it is a circle with center at (0, 0) and radius 5.(10 votes)

- Is a line that crosses a circle -that is on the origin- always have points of intersection of (x,y) and (-x,-y)?(3 votes)
- It should cause lines are lines that goes for ever but it wont if its a graph of a Piecewise Function ( a graph of one is like limits on a line more like a ray. )

https://www.khanacademy.org/math/algebra/algebra-functions/piecewise-functions/v/piecewise-function-example(3 votes)

- I would like to know how to solve, hence to plot these types of equation by the graphical method!? I have been looking all over for the way but with no success! For example x^2 + y^2 = 25, so I know by knowledge that, that's a circle with the four points taken from the square root of 25 but I also want to know how to plot this if let's say, I didn't know that fact. For example how do I solve x(x + 2y) = -4 and 2x + y = 2. Also, does x(x + 2y) = -4 simplify to x^2 + 2xy = -4? Im really stuck with this!(2 votes)
- Yes the x(x + 2y) = -4 does simplify to x^2 + 2xy = -4. You will always know the fact that it is four points from x^2+y^2 .(4 votes)

## Video transcript

What are the solutions to
the system of equations y is equal to x plus
1, and x squared plus y squared is equal to 25? So let's first just visualize
what we're trying to do. So let me try to roughly
graph these two equations. So my y-axis, this is my x-axis. This right over here, x
squared plus y squared is equal to 25, that's going
to be a circle centered at 0 with radius 5. You don't have to know
that to solve this problem, but it helps to visualize it. So if this is 5,
this is 5, 5, 5. This right over
here is negative 5. This right over
here is negative 5. This equation would
be represented by this set of
points, or this is a set of points that
satisfy this equation. So let me-- there you go. Trying to draw it as close
to a perfect circle as I can. And then y equals
x plus 1 is a line of slope 1 with a 1 y-intercept. So this is 1, 2, 3, 4. y-intercept is there
and has a slope of 1 so it looks something like this. So when we're looking
for the solutions, we're looking for the
points that satisfy both. The points that satisfy both
are the points that sit on both. So it's that point-- let me do
it in green-- It's this point and it's this point
right over here. So how do we actually
figure that out? Well, the easiest way
is to-- well, sometimes the easiest way is to substitute
one of these constraints into the other constraint. And since they've already
solved for y here, we can substitute y in the
blue equation with x plus 1 with this constraint
right over here. So instead of saying x squared
plus y squared equals 25, we can say x squared plus,
and instead of writing a y, we're adding the constraint
that y must be x plus 1. So x squared plus x plus 1
squared must be equal to 25. And now, we can
attempt to solve for x. So we get x squared
plus-- now we square this. We'll get-- let me
write in magenta-- we'll get x squared
plus 2x plus 1, and that must be equal to 25. We have 2x squared--
now, I'm just combining these two
terms-- 2x squared plus 2x plus 1 is equal to 25. Now, we could just use
the quadratic formula to find the-- well,
we have to be careful. We have to set this
equal to 0, and then use the quadratic formula. So let's subtract
25 from both sides, and you could get 2x squared
plus 2x minus 24 is equal to 0. And actually, let's--
just to simplify this-- let's divide both sides by 2,
and you get x squared plus x minus 12 is equal to 0. And actually, we don't even have
to use the quadratic formula, we can factor this
right over here. What are two numbers that
when we take their product we get negative 12, and when
we add them, we get positive 1? Well, positive 4 and negative
3 would do the trick. So we have x plus 4 times
x minus 3 is equal to 0. So x could be equal
to-- well, if x plus 4 is 0 then that would make
this whole thing true. So x could be equal
to negative 4 or x could be equal to positive 3. So this right over
here is a situation where x is negative 4. This right over here is
a situation where x is 3. So we're almost done, we just
have to find the corresponding y's. And for that, we can just
resort to the simplest equation right over here, y is x plus 1. So in this situation
when x is negative 4, y is going to be that plus 1. So y is going to be negative 3. This is the point negative
4 comma negative 3. Likewise, when x is 3, y
is going to be equal to 4. So this is the point 3 comma 4. These are the two solutions
to this non-linear system of equations.