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Math Help - points of intersection

  1. #1
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    points of intersection

    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.
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  2. #2
    Super Member ILikeSerena's Avatar
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    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.
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  3. #3
    MHF Contributor MarkFL's Avatar
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    Re: points of intersection

    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)
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  4. #4
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    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

    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?
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  5. #5
    MHF Contributor Also sprach Zarathustra's Avatar
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    Re: points of intersection

    Quote Originally Posted by skoker View Post
    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
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  6. #6
    Super Member ILikeSerena's Avatar
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    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.
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  7. #7
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    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 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.
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  8. #8
    Super Member ILikeSerena's Avatar
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    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 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.
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  9. #9
    MHF Contributor MarkFL's Avatar
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    Re: points of intersection

    Sorry for the omission of the back-substitution...I felt it was evident enough not to mention.
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  10. #10
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    Re: points of intersection

    Quote Originally Posted by ILikeSerena View Post
    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*}
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  11. #11
    Super Member ILikeSerena's Avatar
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    Re: points of intersection

    Quote Originally Posted by Prove It View Post
    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.
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  12. #12
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    Re: points of intersection

    Quote Originally Posted by ILikeSerena View Post
    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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