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How to solve for x? tricky algebra equation

  1. Sep 23, 2011 #1


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    How do you go about solving equations for x that look something like this:

    x3 - x2 + x - 2 = 0

    I'm stumped when I get to:

    x(x2 - x + 1) = 2

    All insights appreciated! :)
  2. jcsd
  3. Sep 23, 2011 #2


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    What kind or level of Algebra? Are you studying College Algebra?

    Try to factor the original polynomial. Can you divide it by (x-1), or (x+1), or (x-2), or (x+2)? Do any of those quotients have no remainder? If you can factor the polynomial in this way, then you may have something in the form, (x-a)(x-b)(x-c)=0, in which you can easily identify the solutions.
  4. Sep 23, 2011 #3


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    I'm actually just taking high school level AP Calc AB, but I was curious how you would solve equations like this one.
  5. Sep 23, 2011 #4


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    This doesn't do you any good at all. Knowing that the product of two numbers is 2 doesn't give you any knowledge about the two numbers, since there are infinitely many pairs of numbers whose product is 2.

    What does do some good is getting a product of factors on one side of the equation, and 0 on the other. If two numbers multiply to zero, then one of the numbers has to be zero.

    What can sometimes help is the Rational Root Theorem. Do a search on wikipedia to find out what that is. I tried all four possibilities for your polynomial and none of them worked. There are some formulas for solving cubics, but they're pretty complicated.

    If all else fails, there are approximation techniques for finding the roots of a polynomial. One such method is Newton's Method, and there are several others.
  6. Sep 23, 2011 #5
    Newton's Method is useless without a graph.
  7. Sep 23, 2011 #6


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    this may be a difficulty with course selection and placement in high school curricula:
    The high school should not simply let you skip Mathematical Analysis or Precalculus. Fortunately when you get to college, you will take either Precalculus or College Algebra, in which you will study various topics about polynomial functions.
  8. Sep 24, 2011 #7


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    I've taken 3 years of algebra, and took pre-calculus last year. I've just forgotten how to do the problem.
  9. Sep 24, 2011 #8


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    As others have already pointed out, that equation is a difficult one as it doesn't factorize.

    Why don't you try an easier one. Say if I change your equation slightly so that it's amenable to factorization. Can you solve this equation?

    x3 - x2 - x - 2 = 0
    Last edited: Sep 24, 2011
  10. Sep 24, 2011 #9


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    I can tell you this about your equation as a function: It has no minimum and no maximum, and it has only one Real zero. Wondering how I know? Graphing calculator.
    Either not factorable or not easily factorable.
  11. Sep 24, 2011 #10


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  12. Sep 24, 2011 #11
    You can focus on x2 - x + 1 to see an approach

    you probably know
    (x-1)2 = x2 - x - x + 1 = x2 - 2*x + 1

    So, what added to itself would equal 1 instead of 2 to make x2 - x + C and how would you get it there. If you find out how to do this, you can figure out the rest.
  13. Sep 24, 2011 #12


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    You should start from the "rational root" theorem: any rational number root of the polynomial equation, [itex]a_nx^n+ a_{n-1}x^{n-1}+ \cdot\cdot\cdot+ a_1x+ a_0= 0[/itex] must be of the form [itex]\frac{m}{n}[/itex] where m evenly divides the constant term [itex]a_0[/itex] and n evenly divides the leading coefficient, [itex]a_n[/itex]. In this problem, the constant term is -2 which has divisors [itex]\pm 1[/itex] and [itex]\pm 2[/itex]. The leading coefficient is 1 which has divisors [itex]\pm 1[/itex]. That is, the only possible rational roots are [itex]\pm 1[/itex] and [itex]\pm 2[/itex]. It easy to see that none of those four possibilities does, in fact, satisfy the equation so it has no rational roots.

    There is, however, "cubic formula, published by Cardano in 1545.

    Note that
    [tex](a- b)^3= a^2- 3a^2b+ 3ab^2- b^3[/tex]
    [tex]3ab(a- b)= 3a^2b- 3ab^3[/tex]

    Adding those [itex](a- b)^3+ 3ab(a- b)= a^3- b^3[/itex]
    If we let x= a- b, m= 3ab, and [itex]n= a^2- b^3[/itex], we see that x= a- b satisfies the "reduced" cubic equation [itex]x^4+ mx= n[/itex] ("reduced" because there is no "[itex]x^2[/itex]" term).

    The question is, "Suppose we know m and n. Can we recover a and b and so find x= a- b?" The answer is "Yes, we can!"

    From m= 3ab, we have b= m/3a. Putting that into [itex]a^3-b^3= n[/itex], [itex]a^3- m^3/(3^3a^3)= n[/itex]. Multiply through by a^3 and we have [itex](a^3)^2- (m/3)^3= na^3[/itex] or [itex](a^3)^2- n a^3- (m/3)^3= 0[/itex], a quadratic equation in [itex]a^3[/itex]. By the quadratic equation,
    [tex]a^3= \frac{n\pm\sqrt{n^2+ 4\left(\frac{m}{3}\right)^3}}{2}= \frac{n}{2}\pm\sqrt{\left(\frac{n}{2}\right)^2+ \left(\frac{m}{3}\right)^3}[/tex]

    Since [itex]a^3- b^3= n[/itex], [itex]b^3= a^3- n= -\frac{n}{2}\pm\sqrt{\left(\frac{n}{2}\right)^2+ \left(\frac{m}{3}\right)^3}[/itex].

    To use that formula on this equation, [itex]x^3- x^2+ x- 2= 0[/itex], first "reduce" it. Let [itex]y= x+ \alpha[/itex] so that [itex]x= y- \alpha[/itex], where [itex]\alpha[/itex] is yet to be determined.

    [tex]x^3= y^3- \alpha y^2+ \alpha^2y- \alpha^3[/tex]
    [tex]-x^2= -y^2+ 2\alpha y- \alpha^2[/tex]
    [tex]+x = y- \alpha[/tex]
    so that [itex]x^3- x^2+ x- 2= y^3- (\alpha+ 1)y^2+ (\alpha^2+ 2\alpha+ 1)y- (\alpha^3- \alpha^2+ \alpha+ 2)[/itex]

    The coefficient of [itex]y^2[/itex], [itex]-(\alpha+ 1)[/itex] will be 0 if [itex]\alpha= -1[/itex].
    And in that case, the coefficient of x, [itex]\alpha^2+ 2\alpha+1= 1- 2+ 1i= 0[/itex] also, and the constant term, [itex]-(\alpha^2+ \alpha+ 2)= -(1- 1+ 2)= -2[/itex]. That is, the equation is equivalent to [itex]y^3- 2= 0[/itex] or [itex]y^3= 2[/itex] so that we don't really need the cubic formula. The roots are the three complex cube roots of 2: [itex]2^{1/3}[/itex], [itex]2^{1/3}(-1/2+ \sqrt{3}i/2)[/itex], and [itex]2^{1/3}(-1/2- \sqrt{3}i/2)[/itex].

    Since [itex]x= y- \alpha[/itex] and [itex]\alpha= -1[/itex], the roots to the original equation are [itex]2^{1/3}+ 1[/itex], [itex]2^{1/3}(1/2+ \sqrt{3}i/2)[/itex], and [itex]2^{1/3}(1/2- \sqrt{3}i/2)[/itex].

    There exist a similarly complicated formula for fourth roots but it can be shown that there cannot be a general formula, using roots (there can be formulas with specially defined functions) or polynomial equations of degree five or higher.
    Last edited by a moderator: Sep 24, 2011
  14. Sep 24, 2011 #13
    You have to be careful when you pick random polynomials of degree higher than 2. You can easily give yourself a very hard problem.

    Here's the solution, as computed by Wolfram Alpha:[PLAIN]http://www3.wolframalpha.com/Calculate/MSP/MSP127319h9270497i48ggd000068ih6e0d8gg59b01?MSPStoreType=image/gif&s=31&w=500&h=62 [Broken]

    Not exactly something I would want to work out by hand...
    Last edited by a moderator: May 5, 2017
  15. Sep 25, 2011 #14
    This is an advanced Algebra problem requiring knowledge of the Factor & Remainder Theorems, Division Algorithm, Zero Factor Property
    Some choices to proceed .............
    1/ Cardano has a formula for 3rd degree polynomials which is similar but more complex than the Quadratic Formula for 2nd degree polynomials
    2/ Newton's Method can find the first root quickly. Then long division to get the remaining quadratic.
    3/ Trial and Error to find the first root often works.
    4/ Rational Root Theorem
  16. Sep 25, 2011 #15

    Char. Limit

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    So draw a graph. Or get out a graphing calculator. It's not that hard.
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