Pre-Calculus - substitution

[elmer]

The x^3 is throwing me off!
Studying Linear and Nonlinear Systems of Equations:

Solve the system by method of substitution

y = -x

y = x^3 + 3x^2 + 2x

I started by substituting -x for y -x = x^3 + 3x^2 + 2x x^3 + 3x^2 + 3x = 0

then my next step was to group (x^3 + 3x^2) + 3x = 0 distribute - x^2(x - 3) + 3x = 0

Would my next step be (x^2 + 3x)(x-3) Help!

 

[DrSteve]

\(\displaystyle -x = x^3 + 3x^2 + 2x\)

So \(\displaystyle x^3 + 3x^2 + 3x=0\)

\(\displaystyle x(x^2 + 3x + 3)=0\)

So \(\displaystyle x=0\) is a solution, and you get two more solutions by solving the quadratic equation \(\displaystyle x^2 + 3x + 3=0\)

After you find these x-values, substitute back into \(\displaystyle y=-x\) to get the corresponding y-values.

 

[lookagain]

\(\displaystyle -x = x^3 + 3x^2 + 2x\)

So \(\displaystyle x^3 + 3x^2 + 3x=0\)

\(\displaystyle x(x^2 + 3x + 3)=0\)

So \(\displaystyle x=0\) is a solution, and you get two more solutions by solving the quadratic equation \(\displaystyle x^2 + 3x + 3=0\)

After you find these x-values, substitute back into \(\displaystyle y=-x\) to get the corresponding y-values.

elmer,

you have the *potential* of getting two more real solutions by solving \(\displaystyle x^2 + 3x + 3 = 0,\)
but because the discriminant is negative, there are no more real solutions.

Then, the only solution of this system of equations is \(\displaystyle (0, 0).\) That is the only place where the graphs
of the curves intersect each other.

 

[DrSteve]

you have the *potential* of getting two more real solutions by solving \(\displaystyle x^2 + 3x + 3 = 0,\)
but because the discriminant is negative, there are no more real solutions.

This statement is only true if we restrict to the real numbers. If we allow complex solutions, then an nth degree polynomial has exactly n roots (counting multiplicity). In particular, every quadratic equation has exactly 2 roots over the complexes (assuming that we count a double root twice - this makes sense because the corresponding factor will appear twice).

 

[lookagain]

\(\displaystyle -x = x^3 + 3x^2 + 2x\)

So \(\displaystyle x^3 + 3x^2 + 3x=0\)

\(\displaystyle x(x^2 + 3x + 3)=0\)

So \(\displaystyle x=0\) is a solution, and you get two more solutions by solving the quadratic equation \(\displaystyle x^2 + 3x + 3=0\)

After you find these x-values, substitute back into \(\displaystyle y=-x\) to get the corresponding y-values.

Only those x-values which are real solutions will be substituted back into \(\displaystyle y = -x,\) to get
the corresponding real y-values, because this problem is to solve for the coordinates of points
in the real xy-plane where the graphs intersect. Imaginary values of x and y are not even considered,
regardless of there being n roots.

 

[Deleted member 4993]

....... because this problem is to solve for the coordinates of points
in the real xy-plane where the graphs intersect. Imaginary values of x and y are not even considered,
regardless of there being n roots.

The original post...

Solve the system by method of substitution

y = -x

y = x^3 + 3x^2 + 2x

Nothing is explicitly stated about "real" solution.

 

[Denis]

Solve the system by method of substitution
y = -x
y = x^3 + 3x^2 + 2x

Why the heck didn't the "teacher/book" simply state:
x^3 + 3x^2 + 3x = 0 ; solve for x

 

[lookagain]

....... because this problem is to solve for the coordinates of points
in the real xy-plane where the graphs intersect. Imaginary values of x and y are not even considered,
regardless of there being n roots.

The original post...

Solve the system by method of substitution

y = -x

y = x^3 + 3x^2 + 2x

\(\displaystyle > \ >\)Nothing is explicitly stated about "real" solution. \(\displaystyle < \ <\)

It doesn't have to be; this is a problem that asks for solutions (read: coordinates) on the
xy-plane, because they correspond to curves intersecting. Otherwise, it wouldn't have had
the equations for the curves.

And -x = x^3 + 3x^2 + 2x (without the y variable) doesn't have that understood. All real
solutions should be solved for in that case.

 

[DrSteve]

I don't see anything in the way this problem is phrased that suggests only real solutions should be found. Since Complex Numbers are generally covered in a precalculus course I would guess that they want all complex solutions.

Since the question isn't specific I would consider both answers correct.

 

[lookagain]

I don't see anything in the way this problem is phrased that suggests only real solutions should be found. .

You can ignore all my posts about it if you want, but that doesn't change the fact that when
the equations of these two curves are to be simultaneously solved for, \(\displaystyle a \ solver \ only \ cares \ about\)
\(\displaystyle where \ they \ intersect \ in \ the \ xy \small{-} \ plane, \ not \ imaginary \ intersections.\)

Else, the onus should rest with the problem maker to alter the nature of the problem.

I don't see anything...

I see nothing!

People see what they want to see.