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Old Jun14-09, 05:45 AM                  #1
jazduck

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Quick Algebra Problem

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)?
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Old Jun14-09, 06:14 AM                  #2
Cyosis

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Re: Quick Algebra Problem

This should be in the homework section. You can solve it by using the ABC-formula.
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Old Jun14-09, 06:32 AM                  #3
jazduck

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Re: Quick Algebra Problem

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.
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Old Jun14-09, 09:56 AM                  #4
Cyosis

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Re: Quick Algebra Problem

The solution to an equation of the form LaTeX Code: ax^2+bx+c=0 is given by LaTeX Code: x=\\frac{-b \\pm \\sqrt{b^2-4ac}}{2a} . To make use of this formula you will have to write LaTeX Code: t^2+t=5.1 as LaTeX Code: t^2+t-5.1=0 .
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Old Jun14-09, 10:08 AM                  #5
HallsofIvy

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Re: Quick Algebra Problem

Alternatively, you could "complete the square".

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

Now that the left side is a "perfect square" you have LaTeX Code: (t+ 1/2)^2= 5.35 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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