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## Quadratic systems

Current time:0:00Total duration:4:14

# Quadratic systems: a line and a circle

CCSS Math: HSA.REI.C.7, HSA.REI.C

## 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.