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Difficulty solving the following quadratic equation

  1. Apr 5, 2015 #1
    < Mentor Note -- thread moved to HH from the technical math forums, so no HH Template is shown >

    Hi,
    I'm having difficulty solving the following quadratic equation

    x^2-(660/x-5)-91=0. The fact that the middle term has an x-5 rather than just an x that is throwing me

    Can anyone please help?
    Kind Regards
    Daniel
     
    Last edited by a moderator: Apr 5, 2015
  2. jcsd
  3. Apr 5, 2015 #2

    phinds

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    A quadratic equation is one that is of the form ax^2 + bx + c. This equation does not meet that criteria. Do you see why?
     
  4. Apr 5, 2015 #3

    robphy

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    Can you do some algebra to rewrite your equation to be in the form ##Ax^2+Bx+C=0## where A, B, and C are constants?
     
  5. Apr 5, 2015 #4
    Yes phinds I do but then how do I solve this eqn?
     
  6. Apr 5, 2015 #5
    I don't see how?
     
  7. Apr 5, 2015 #6

    FactChecker

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    There might be a trick that makes this easier, but I don't see it. You can always multiply through by the denominator x (or is x-5 the denominator?) and expand to Ax^3 + Bx^2 + Cx + D = 0. Those have closed form solutions, but they involve some complex arithmetic and a re more difficult than solving a quadratic.
     
  8. Apr 5, 2015 #7

    symbolipoint

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    Your equation is RATIONAL; not quadratic. The denominator is x, as written. Condense to a single rational expression and look for the zeros.
     
  9. Apr 5, 2015 #8

    SammyS

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    No, this cannot be done for the equation in question.


    If your equation is ##\displaystyle\ x^2-\frac{660}{x-5}-91=0\,,\ ## then you should have included the extra parentheses as follows below.

    x^2-(660/(x-5))-91=0.
     
    Last edited: Apr 5, 2015
  10. Apr 5, 2015 #9
    Can you expand on this symbolipoint? If I convert it to a single expression it will become a cubic equation where it will still be problematic to resolve all roots
     
  11. Apr 5, 2015 #10

    symbolipoint

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    It will be still a RATIONAL equation, not a cubic equation. The numerator will be cubic but the denominator will be linear.

    x^2-(660/x-5)-91=0
    x^2-660/x+5-91=0
    x^2-660/x-86=0
    x^2(x/x)-660/x-86(x/x)=0
    (x^3-660-86x)/x=0
    (x^3-86x-660)/x=0
     
  12. Apr 6, 2015 #11
    apologies but my equation is actually x^2-(660/(x-5))-91=0 so that solution will not work
     
  13. Apr 6, 2015 #12

    ehild

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    Multiply the whole equation with (x-5): you get a cubic equation. Try to find a simple root among the dividers of the constant term. (There is one)
     
  14. Apr 6, 2015 #13

    statdad

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    Not if you clear the left of all fractions - that was the point behind the "cubic equation" comment above.
     
  15. Apr 6, 2015 #14

    symbolipoint

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    That changes the expression, and therefore the function. The task of checking for the ROOTS would be equivalent to dealing with the properly resulting quadratic equation. Objective finally understood.
     
  16. Apr 7, 2015 #15

    statdad

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    There is no "function" here, just an expression in an equation.
     
  17. Apr 7, 2015 #16

    SammyS

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    There is an equation to be solved. The left hand side of the equation is a function.

    One well accepted way to solve the equation is to multiply both sides by the denominator appearing on the on the left hand side, namely (x-5). The result is an equation with a cubic function on the right hand side, i.e. a cubic equation. Solving this equation is equivalent to finding the zeros of that cubic function.

    Another well accepted way to solve the original equation is to express the left hand side formally as a rational function by using (x-5) as a common denominator.
    ##\displaystyle\ x^2-\frac{660}{x-5}-91\ ## becomes ##\displaystyle\ \frac{(x^2-91)(x-5)-660}{x-5}\ ##
    The zeros of this function are the same as the zeros of the numerator. The numerator is the same cubic function found in the first method of solution.

    As pointed out by ehild in Post #12, there is one easy to find root for this cubic equation.

    The other two roots can then be found in a straight forward manner.
     
  18. Apr 7, 2015 #17

    statdad

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    "The left hand side of the equation is a function."

    No, there is a rational expression there, as the equation is written. I understand all the work to find solutions to the equation.
     
  19. Apr 7, 2015 #18

    SammyS

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    What is it that makes the left hand side not be a function?
     
  20. Apr 7, 2015 #19

    symbolipoint

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    That is the basic method described for resolving the whole problem.
     
  21. Apr 14, 2015 #20

    statdad

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    Usage. The left side of that equation is a rational expression - it is not considered in isolation.
     
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