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Rearranging a formula

  1. Nov 4, 2016 #1
    1. Rearrange the following equation
    2/(x-2) + 2/(x+2) = 1/2
    into the form Ax2+ Bx + C = 0

    I did it and got -x2 + 8x + 4 = 0

    Am I correct please?
     
  2. jcsd
  3. Nov 4, 2016 #2

    berkeman

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    Could you please show your step-by-step work? That will make it easier for us to check it.
     
  4. Nov 4, 2016 #3

    FactChecker

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    Looks good to me. You should usually show your work in step-by-step fashion. Trying to do too much in your head is slow and error-prone.
     
  5. Nov 4, 2016 #4
    See picture attached for working out
     

    Attached Files:

  6. Nov 4, 2016 #5

    Ray Vickson

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    No, your final equation in incorrect. Start again, and this time do it the easy way: re-write ##2/(x-2) + 2/(x+2)## by putting both terms over a common denominator. I really could not figure out what you were trying to do in your original working, but whatever it was produced errors.
     
  7. Nov 4, 2016 #6
    Where am I going wrong... I need to rearrange the following equation 2/(x-2) + 2/(x+2) = 1/2 into the form Ax2+ Bx + C = 0

    Here's my work:

    2/(x-2) + 2/(x+2) = 1/2

    2 + 2(x-2)/(x+2) - (x-2)/2 = 0
    2(x+2) + 2(x-2) - (x-2)(x+2)/2 = 0
    4(x+2)+4(x-2)-(x-2)(x+2) = 0
    4x+8+4x-8-(x2 - 2x+2x-4) = 0
    8x-x2+4 = 0
    -x2 +8x+4 = 0
     
  8. Nov 4, 2016 #7

    FactChecker

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    I think that your answer is correct ... with two caveats.
    1) Always keep track of values of x that are invalid: x=2 or x=-2 cause a divide by 0 in the original equation. So always add the constraints x≠2; x≠-2. (Click on the 'Σ' button at the top of the entry window to get the '≠' symbol, among others.)
    2) Use clear notation for the exponent 2. If you do not have superscripts, type x^2. (On the web site, click on the 'x2' button at the top of the entry window to get superscripts.)

    So your final equation should look like:
    -x2 +8x+4 = 0; x ≠ +2; x ≠ -2
     
  9. Nov 4, 2016 #8

    Ray Vickson

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    Sorry: I missed the minus sign in front of your x^2 term, so ##x^2 -8x - 4=0## IS correct. You wrote it as ##-x^2 + 8x + 4 = 0##, but the usual way of writing polynomial equations is to have the coefficient of the highest power of ##x## being positive (so with ##+x^2## instead of ##-x^2##).
     
  10. Nov 4, 2016 #9
    I'm sorry thank you for the advice
     
  11. Nov 4, 2016 #10
    Why is it then I get two different solution for both equations -x^2 + 8x +4 = 0 and x^2 -8x - 4 = 0 ?
     
  12. Nov 4, 2016 #11

    Ray Vickson

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    They are not two different equations! If ##x^2 -8x-4=0## then also ##0 = -0 = -(8x^2 -8x-4) = -x^2 + 8x + 4##.

    OK, I guess you could say they are different, just as ##5x^2 - 40 x - 20=0## looks different; but they are all just simple multiples of each other, and they all have exactly the same solutions.
     
    Last edited: Nov 5, 2016
  13. Nov 6, 2016 #12

    SammyS

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    @Natasha1 ,

    Thus they are called equivalent equations .
     
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