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Quick Algebra Problem

  1. Jun 14, 2009 #1
    I have an equation which I ended up with as t^2 + t = 5.1

    How do I then solve for t (without just plugging it into solve on the calculator)?
     
  2. jcsd
  3. Jun 14, 2009 #2

    Cyosis

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    This should be in the homework section. You can solve it by using the ABC-formula.
     
  4. Jun 14, 2009 #3
    Cyosis - The question is not homework so I put it in general math. The actual question i'm trying is one on the initial velocity of an object required to catch another object released a second before it over a 50m displacement. Not that hard if I make my calculator solve it, just making a mess of the algebra because I havn't done it in 8 years.

    Anyway will look up the ABC rule.
     
  5. Jun 14, 2009 #4

    Cyosis

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    The solution to an equation of the form [itex]ax^2+bx+c=0[/itex] is given by [itex]x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}[/itex]. To make use of this formula you will have to write [itex]t^2+t=5.1[/itex] as [itex]t^2+t-5.1=0[/itex].
     
  6. Jun 14, 2009 #5

    HallsofIvy

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    Alternatively, you could "complete the square".

    For any number, a,
    [tex](t- a)^2= t^2- 2at+ a^2[/tex]
    You have [itex]t^2+ t= 5.1[/itex] which will match the first two terms of that if 2at= t or a= 1/2. In that case [itex]a^2= 1/4[/itex] so adding 1/4 to both sides of the equation: [itex]t^2+ t+ 1/4= 5.1+ 1/4= 5.1+ .25= 5.35[/itex]

    Now that the left side is a "perfect square" you have [itex](t+ 1/2)^2= 5.35[/itex] and you can solve that by taking the square root of both sides (remembering that the result can be either positive or negative).
     
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