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Simple Algebra Problem

  1. Jan 3, 2006 #1
    Ok, I have a simple algebra problem I need help with.

    The problem states "The product of two consecutive even integers is 48."
    So I wrote the equation out as x(x+2)=48. If I factor that I should get x^2+2x=48? Then divide 48 by 2x & get 24. Then squard root of 24? What step am I doing incorrect? I've been racking my brain on this problem for like a half an hour :tongue: (this is what I get for being out of school for 4 years). I mean the answer is simple 6 x 8 = 48. But, I would like to know the steps. Any help is appreciated.
     
    Last edited: Jan 3, 2006
  2. jcsd
  3. Jan 3, 2006 #2
    ok so you have
    [tex] x(x+2) = 48 [/tex]
    [tex] x^2 +2x = 48[/tex]

    The thing is that you can't just divide by 2x and get [tex] x^2 = 24[/tex]
    rewrite the equation as
    [tex] x^2 + 2x - 48 = 0[/tex]

    then use the quadratic equation to solve for x, and take the solution that fits the problem's description.
     
    Last edited by a moderator: Jan 3, 2006
  4. Jan 3, 2006 #3
    Ok, brilliant. Thanks much for your help. It is greatly appreciated :cool:. The algebra rules are slowly coming back to me.
     
  5. Jan 4, 2006 #4
    Yeah, if you divide [tex]x^2 + 2x = 48[/tex] by 2x then you get:

    [tex]\frac {1}{2}x + 1 = \frac {24}{x}[/tex]

    Doesn't really get you anywhere. You have to factor to:
    [tex](x-6)(x+8) = 0[/tex]
     
  6. Jan 5, 2006 #5
    One more way. x^2+2x=(x+1)^2-1. So, (x+1)^2=49 So, x+1=(+/-)7
     
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