Re: points of intersection

Hi skoker! :)

It seems to me you have the right approach.

Raise both sides of your equation to the 3rd power and you get a 6th order polynomial.

It has 2 obvious roots which are the only real roots.

Re: points of intersection

I would go with your first approach, giving:

Let and write:

Thus, we get 2 points of intersection:

(-1,-1), (1,1)

Re: points of intersection

i guess it is the only way to raise the cube.

@MarkFL2

this substitution you did, what is it called?

if you use the normal algebra and raise both sides you get

then you start the factor.

but what you did there looks like the U-substitution from calculus for finding the anti-derivative of a composite function.

where in the algebra is this type of substitution explained? maybe for parametric equations?

Re: points of intersection

Quote:

Originally Posted by

**skoker** i guess it is the only way to raise the cube.

@MarkFL2

this substitution you did, what is it called?

if you use the normal algebra and raise both sides you get

then you start the factor.

but what you did there looks like the U-substitution from calculus for finding the anti-derivative of a composite function.

Read here:

Rational Zeros of Polynomials

Re: points of intersection

It is called "substitution".

Just what you already said.

If you have an equation with a difficult sub expression, you give the sub expression a name, say "u", and replace it.

Then you solve the equation, and when you're done, you back substitute "u".

Actually, substitution in integrals is a more advanced version of regular substitution.

Re: points of intersection

thanks for the info...

@ILikeSerena

i will go back and look at the substitution a bit. it seems in the algebra they only give some easy examples.

like substitute for

for factoring. but not how to use with roots and more complicated expressions.

until they show how to do the u-substitution.

Re: points of intersection

Well, MarkFL2 left out the back substitution.

Basically he finished with the roots u=1 and u=-1.

Since he defined , back substitution means you still have to solve:

and

.

The solutions for those are x=1 and x=-1.

Re: points of intersection

Sorry for the omission of the back-substitution...I felt it was evident enough not to mention. :)

Re: points of intersection

Quote:

Originally Posted by

**ILikeSerena** Well, MarkFL2 left out the back substitution.

Basically he finished with the roots u=1 and u=-1.

Since he defined

, back substitution means you still have to solve:

and

.

The solutions for those are x=1 and x=-1.

As well as

Re: points of intersection

Quote:

Originally Posted by

**Prove It** As well as

I have to admit that I assumed that the points where the 2 graphs intersect have real coordinates.

Otherwise, we would also need to consider the imaginary solutions for u. ;)

Re: points of intersection

Quote:

Originally Posted by

**ILikeSerena** I have to admit that I assumed that the points where the 2 graphs intersect have real coordinates.

Otherwise, we would also need to consider the imaginary solutions for u. ;)

You're not thinking fourth dimensionally ;) or even third, for that matter :P hahaha.