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Have i done this right?

  1. Dec 5, 2008 #1
    1. The problem statement, all variables and given/known data

    Needing to factor the relevent equation.

    2. Relevant equations

    (pM)= ½ (F²vt/c)+ ½ (F²vt/v)

    Square both sides:

    (pM)²= ½ (F²vt/c)² + ½ (F²vt/v)²

    3. The attempt at a solution

    (pM)= ½ (F²vt/c)+ ½ (F²vt/v)

    Square both sides:

    (pM)²= ½ (F²vt/c)² + ½ (F²vt/v)²

    Then factor and simplify

    ½ F²(vt/c – vt/v)= p²M²
     
  2. jcsd
  3. Dec 5, 2008 #2
    If I square both sides of [tex]y = \frac{1}{2}x[/tex], I don't get [tex]y^2 = \frac{1}{2}x^2[/tex] ... I get [tex]y^2 = \frac{1}{4}x^2[/tex].

    If I square both sides of [tex]y = a + b[/tex], I don't get [tex]y^2 = a^2 + b^2[/tex] ... I get [tex]y^2 = (a+b)^2 = a^2 + 2ab + b^2[/tex].
     
  4. Dec 5, 2008 #3
    And might I also suggest that factoring a bit earlier might simplify the problem.
     
  5. Dec 5, 2008 #4

    In other words i have squared wrongly... which i have... hold on, because one i properly do this, i still need to factor it out correctly.
     
  6. Dec 5, 2008 #5
    How would i do that exactly?

    Thanks by the way.

    I ask because i am bit confused about my operations.
     
  7. Dec 5, 2008 #6
    Well, if I had an equation of the form [tex]y = abx - abt[/tex], and wanted to solve for x, I might want to consolidate the common factors first:

    [tex]y = ab (x - t)[/tex]

    Then move them over to the other side by dividing through by those common factors:

    [tex]\frac{y}{ab} = x - t[/tex]

    Hmm. Only one step remains, and it can be done by adding an element to both sides.
     
  8. Dec 5, 2008 #7

    But concerning my equations... Is this relevent, because i am a terrible problem solver
     
  9. Dec 5, 2008 #8
    What do these fellows have in common?

    ½ (F²vt/c)+ ½ (F²vt/v)

    And I hate to ask this late in the game, but what are we solving for, here? Do we stop at factoring, or go on to solve for c or v? Because the step above is as good a run at factoring as you're likely to get. The rest is algebra (adding, subtracting, multiplying, dividing, etc through the entire equation).
     
  10. Dec 5, 2008 #9
    We are factoring out the F. And we also solve for F... if that makes sense? If it doesn't, then perhaps i need to solve for v or c?
     
  11. Dec 5, 2008 #10
    Yep, F's easy. But try to figure out the problem yourself first. Hint:

    If

    [tex]f = \frac{abx}{c} + \frac{abx}{x}[/tex]

    then

    [tex]f = abx(\frac{1}{c} - \frac{1}{x})[/tex]

    or equivalently (if we leave the x behind)

    [tex]f = ab(\frac{x}{c} - 1)[/tex]
     
  12. Dec 5, 2008 #11
    Right, o'k. I can work that out. Hold on. Cheers.
     
  13. Dec 5, 2008 #12

    pM=Fvt/c+Fvt/v

    pM=Fv(t/c– 1)

    Is the only pattern I see in this. Is this formula supposed to help, because I am really confused. This is the second time I have came here, funnily enough to be such confused again. But if ya could help, I’d appreciate a more direct response, because I am absolutely no good at mathematical riddles, or anything for me to solve. I don’t have a mathematical brain really ;)
     
  14. Dec 5, 2008 #13
    Well, they're not really riddles. See, I could show you a gun and shoot a duck. Then I could hand you a gun and ask you to shoot a crow. The wrong answer in that case would be "But you didn't show me how to shoot a crow, just a duck."

    Teaching by way of example is a time honored tradition. If I say I can solve y = 5x for x by dividing through by 5, I get y/5 = x. Ta daah.

    But what about y = 6x? Same trick. y/6 = x

    But what about y = kx, where k is some constant? Same trick. y/k = x. Ta daah.

    What about a whole conga line of constants? If y = abcdefgx, hey, I can pick em off 1 by 1 like . . . well, like shooting ducks . . . and get y/(abcdefg) = x

    Sometimes we answer in examples. Not riddles.

    Past that? It's homework help, not solutions. I could hand over an answer, and you could repeat it, but . . .

    there will be more ducks.

    Anyways, your factorization left out a number of things. Try this on and see if you can work it from here. Better yet, try to figure out how I came up with this.

    Given:

    (pM)= ½ (F²vt/c)+ ½ (F²vt/v)

    The factors can be combined to this just-about-ready-to-solve-for-F type form:

    [tex]
    2 p M = F^2 t ( \frac{v}{c} - 1 )
    [/tex]
     
  15. Dec 5, 2008 #14
    Simple algebraic rules otherwise, like

    E=Mc^2

    M=E/C^2

    C^2=E/M

    I know this. This is not what i was asking in the first place. But, you say the last part, and i suppose this is the proper way to factor this out for F?
     
  16. Dec 5, 2008 #15
    Good response, now apply it and solve for F^2 as you have solved for C^2 in your example.
     
  17. Dec 5, 2008 #16
    And you can check to see if it's a proper factorization by multiplying it out, reversing the procedure.
     
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