1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

How to solve and equation?

  1. Nov 8, 2007 #1
    [SOLVED] how to solve and equation?

    1. The problem statement, all variables and given/known data
    sqrt(ax^2 +bx) - sqrt(cx^2 + dx) = e
    where you know a,b,c,d and e


    2. Relevant equations
    i dont know


    3. The attempt at a solution
    i put the entire equation on the power of 2 and then simplified adn then repeated again ... and i got an equation of 4th grade (ax^4+bx^3+.......=0)
    and well i cant solve it so i retuned back to my original problem :)
     
  2. jcsd
  3. Nov 8, 2007 #2

    Astronuc

    User Avatar

    Staff: Mentor

    If the first sqrt argument was a form (x+p)2 and the second sqrt argument was (x+q)2, then one would have a form

    x+p + x+q = e

    or x = 1/2 (e - p - q) and don't forget roots +/-
     
  4. Nov 13, 2007 #3
    hmm sadly its not a full square :(

    solved it with taylor series (got and estimate and its better then nothing)
     
  5. Nov 13, 2007 #4
    I'm talking only about the case when a, b, c, d and e are integers.

    First of all, substract e from both sides you get the form (let d - e = f) ax[tex]^{4}[/tex] + bx[tex]^{3}[/tex] + cx[tex]^{2}[/tex] + f = 0

    Then you have to factor the equation to a(x - x[tex]_{1}[/tex])(x - x[tex]_{2}[/tex])(x - x[tex]_{3}[/tex])(x - x[tex]_{4}[/tex]) = 0.

    To do this you must first figure out one of the roots. All roots are in the set of [tex]\frac{d}{a}[/tex] divisors, so what I usually do is try to input all those divisors as x. Now, when you find one root, you must input it in Horner's scheme.

    1)In Horner's scheme you put in the line A all coefficients of x[tex]^{n}[/tex] - a, b, c and d. In position C1 (look at the example) you place the root you guessed. Now rewrite the coefficient from A2 to C2. Multiply C2 with the root (B1) and write the number at B3. Now add to it A3 and write it at C3. Multiply C3 with the root and rewrite it at B4. Add A4 to B4 and rewrite the sum at C4. Now multiply it with the root and rewrite the product at B5. etc. Do you see the algorithm? If no, then here it might me explained a bit better. Anyway, if C6 is not 0 then check your calculations again or the root might be not a root. Now the C line are coefficients for (x - x[tex]_{1}[/tex])(C2x[tex]^{3}[/tex] + C3x[tex]^{2}[/tex] + C4x + C5) = 0. Now factor the part in second brackets until you reach the form previously mentioned where x[tex]_{1}[/tex], x[tex]_{2}[/tex] etc. are the roots you need.

    Here's an example:

    5x[tex]^{4 [/tex] - 3x[tex]^{3}[/tex] - 4x[tex]^{2}[/tex] - 3x + 5 = 0

    I find that one of the roots is x[tex]_{1}[/tex] = 1 so

    1. 2. 3. 4. 5. 6. ==== position numbers
    ---------------------------------
    | 5 -3 -4 -3 5 (A)
    |
    1 | 5 2 -2 -5 (B)
    |-----------------------------
    5 2 -2 -5 0 (C)

    Now I have 5(x - 1)(5x[tex]^{3}[/tex] + 2x[tex]^{2}[/tex] - 2x - 5) = 0

    On 5x[tex]^{3}[/tex] + 2x[tex]^{2}[/tex] - 2x - 5 I use Horner's scheme again. I guess that one root x[tex]_{2}[/tex] = 1 and...

    | 5 2 -2 -5
    |
    1 | 5 7 5
    |--------------------------
    | 5 7 5 0

    Now I have 25(x - 1)(x - 1)(5x[tex]^{2}[/tex] + 7x + 5) = 0

    You probably now how to factor 5x[tex]^{2}[/tex] + 7x + 5 so I get

    125(x - 1)(x - 1)(x - (7 + 2[tex]\sqrt{6}[/tex])/2)(x - (7 - 2[tex]\sqrt{6}[/tex])/2) = 0

    If the result is 0 then at least one of the multipliers must be 0. So x[tex]_{1}[/tex] = x[tex]_{2}[/tex] = 1; x[tex]_{3}[/tex] = (7 + 2[tex]\sqrt{6}[/tex])/2; x[tex]_{4}[/tex] = (7 - 2[tex]\sqrt{6}[/tex])/2).

    I hope that helps.

    Whew, that tok a lot of time. If something is unclear, and it probably will be from what I've written, feel free to ask.
     
  6. Nov 15, 2007 #5
    tried it and it works :D thank you (does it work for x^n+... aswell?)
     
  7. Nov 15, 2007 #6
    Yes, in x[tex]^{n}[/tex]+... n is the max number of roots by the way.
     
  8. Nov 16, 2007 #7
    cool :)

    thanks
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: How to solve and equation?
Loading...