points of intersection

**skoker** · December 22nd 2011, 09:50 PM

i am not sure how to find the points of intersection of these 2 functions algebraically.

$y=x^{1/3}$ and $y=x^2+x-1$

i would normally do this

$x^{1/3}=x^2+x-1$ and solve for x. but in this case the cube is making this a nasty approach.

one try got me to here $0=x^{5/3}+x^{2/3}-2$

but this seems like i have wondered into the weeds.

**ILikeSerena** · December 22nd 2011, 10:00 PM

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.

**MarkFL** · December 22nd 2011, 10:23 PM

I would go with your first approach, giving:

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

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

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

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

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

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

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

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

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

Thus, we get 2 points of intersection:

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

**skoker** · December 23rd 2011, 11:22 AM

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

$x^6+3x^5-5x^3+2x-1$ 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?

**Also sprach Zarathustra** · December 23rd 2011, 11:27 AM

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

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

**ILikeSerena** · December 23rd 2011, 11:33 AM

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.

**skoker** · December 23rd 2011, 12:16 PM

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 $u=x^2$ for $x^4+x^2... = u^2+u...$

for factoring. but not how to use with roots and more complicated expressions.
until they show how to do the u-substitution.

**ILikeSerena** · December 23rd 2011, 12:25 PM

Well, MarkFL2 left out the back substitution.

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

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

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

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

**MarkFL** · December 23rd 2011, 12:43 PM

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

**Prove It** · December 23rd 2011, 04:47 PM

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

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

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

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

**ILikeSerena** · December 23rd 2011, 04:54 PM

Originally Posted by Prove It

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

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.

**Prove It** · December 23rd 2011, 05:00 PM

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.

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