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:

$\displaystyle x^2+x-x^{\small{\frac{1}{3}}}-1=0$

Let $\displaystyle u=x^{\small{\frac{1}{3}}}$ and write:

$\displaystyle u^6+u^3-u-1=0$

$\displaystyle u^3\left(u^3+1\right)-(u+1)=0$

$\displaystyle u^3(u+1)\left(u^2-u+1\right)-(u+1)=0$

$\displaystyle (u+1)\left(u^3\left(u^2-u+1\right)-1\right)=0$

$\displaystyle (u+1)\left(u^4\left(u-1\right)+u^3-1\right)=0$

$\displaystyle (u+1)\left(u^4\left(u-1\right)+(u-1)\left(u^2+u+1\right)\right)=0$

$\displaystyle (u+1)(u-1)\left(u^4+u^2+u+1\right)=0$

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

$\displaystyle x^6+3x^5-5x^3+2x-1$ 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

$\displaystyle x^6+3x^5-5x^3+2x-1$ 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 $\displaystyle u=x^2$ for $\displaystyle x^4+x^2... = u^2+u...$

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 $\displaystyle u=\sqrt[3]x$, back substitution means you still have to solve:

$\displaystyle \sqrt[3]x=1$ and $\displaystyle \sqrt[3]x=-1$.

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 $\displaystyle u=\sqrt[3]x$, back substitution means you still have to solve:

$\displaystyle \sqrt[3]x=1$ and $\displaystyle \sqrt[3]x=-1$.

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

As well as $\displaystyle \displaystyle \begin{align*} x = e^{-\frac{2\pi i}{3}} , e^{-\frac{\pi i}{3}}, e^{\frac{\pi i}{3}}, e^{\frac{2\pi i}{3}} \end{align*}$

Re: points of intersection

Quote:

Originally Posted by

**Prove It** As well as $\displaystyle \displaystyle \begin{align*} x = e^{-\frac{2\pi i}{3}} , e^{-\frac{\pi i}{3}}, e^{\frac{\pi i}{3}}, e^{\frac{2\pi i}{3}} \end{align*}$

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.