Cubic Equations

Discussion in 'Other Pre-University Math' started by nycmathguy, Jul 12, 2021.

  1. nycmathguy

    nycmathguy

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    Set 1.3
    Question 48
    David Cohen

    48. Solve each of the following equations for x. Hint: As in the text, begin by factoring out a common factor.

    (a) x^3 -13x^2 + 42x = 0

    (b) x^3 -6x^2 + x = 0

    The above is from the David Cohen Precalculus textbook. The extra information below is not from the textbook.

    Note: Do not solve 48 (a & b).

    Factoring out the common factor for question 48 (a & b) is the logical thing to do as step one.

    However, what if we were not able to factor out a common factor for a cubic equation?

    Try these two:

    (a) x^3 -13x^2 + 42 = 0

    (b) x^3 -6x^2 +sqrt{2} = 0

    Solve for x.
     
    nycmathguy, Jul 12, 2021
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  2. nycmathguy

    MathLover1

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    Solve for x:

    (a)

    x^3 -13x^2 + 42 = 0....cannot be factored

    you can solve it using Newton- Raphson method
    Newton's Method is based upon finding roots of a function f(x). To see how this applies to square or cube roots, suppose that y=√n for some fixed n. Well, then this y would be a root of the equation f(x)=x^2-n. Similarly, f(x)=x^3-n would provide us with a way to calculate the cube root of n.

    or, you can solve it using The Cubic Formula
    here is that formula:
    https://math.vanderbilt.edu/schectex/courses/cubic/

    as you can see, it is complicated to do
    I suggest leave it for now and try those cubic equations that could be factored


    (b) x^3 -6x^2 +sqrt(2) = 0....cannot be factored either
     
    MathLover1, Jul 13, 2021
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    nycmathguy likes this.
  3. nycmathguy

    nycmathguy

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    It is beyond the scope of precalculus. I get it. Thanks again. I want to see more tutors here. I really miss freemathhelp.com. Very good tutors there.
     
    nycmathguy, Jul 13, 2021
    #3
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