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Polynomial equation

  1. Feb 4, 2004 #1
    i need help with a couple of problems

    1.

    (x+3)^2/3=3

    I don't even know were to start

    2.1 - 1 = 3
    - ---
    x x+1

    LCD is (x)(x+1)

    1(x)(x+1) - 1(x)(x+1) = 3(x)(x+1)
    --------- --------- ---------
    x(x+1) (x)x+1 (x)(x+1)

    1x+1 - 1x = 3x^2 + 3x

    3x^2 + 3x + 1 = 0

    3^2 - 4(3)(1)

    that is all i have and i don't think it is right because the answer should be

    -3 + or - the squ. root of 11
    -----------------------------
    6
     
  2. jcsd
  3. Feb 4, 2004 #2

    NateTG

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    Homework Helper

    You've got:
    [tex]\frac{(x+3)^2}{3}=3[/tex]
    so
    [tex](x+3)^2=9[/tex]
    right?
     
  4. Feb 5, 2004 #3

    HallsofIvy

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    Assuming that (x+3)^2/3=3 meant [tex]\frac{(x+3)^2}{3}=3[/tex] then
    NateTG is correct: (x+3)2= 9.

    2. is a little hard to understand. I think you meant
    [tex]\frac{1}{x(x+1)}-1= 3[/tex].

    Yes, the LCD is x(x+1) but then you used that incorrectly. If you want to combine fractions on the left, the first fraction remains [tex]\frac{1}{x(x+1}[/tex]. You don't want the "x(x+1)" in the numerator. You multiply both numerator and denominator of the "-1" by x(x+1) to get [tex]\frac{1}{x(x+1)}- \frac{x(x+1)}{x(x+1)}= 3[/tex]. The "3" also does not change on the right hand side.

    Actually, rather than adding the fractions I would recommend simply multiplying both sides of the equation by the LCD:
    [tex]x(x+1)(\frac{1}{x(x+1)}-1)= 3(x(x+1))[/tex]so
    [tex]1- x(x+1)= 3x(x+1)[/tex] or
    [tex] 2x(x+1)= 2x^2+ 2x= 1[/tex]
    Now solve that by completing the square or using the quadratic formula.
     
    Last edited: Feb 7, 2004
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