Non-linear systems of equations 3 Video Lecture | Mathematics for Digital SAT

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.