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Homework Help: SAT Math Question

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  1. Nov 1, 2008 #1
    1. The problem statement, all variables and given/known data
    [tex]\sqrt{x^{2}-t^{2}}=2t-x[/tex]

    If x and t are positive numbers that satisfy the equation above, what is the value of [tex]\frac{x}{t}[/tex]?

    2. The attempt at a solution
    [tex]x^{2}-t^{2}=4t^{2}-x^{2}[/tex]

    [tex]x^{2}+x^{2}=4t^{2}+t^{2}[/tex]

    [tex]2x^{2}=5t^{2}[/tex]

    I'm unsure of what to do next.
     
  2. jcsd
  3. Nov 1, 2008 #2

    gabbagabbahey

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    Divide both sides of the equation by [itex]2t^2[/itex] and then take the square root.
     
  4. Nov 1, 2008 #3
    Do you mean divide:

    [tex]2x^{2}=5t^{2}[/tex] by [tex]2t^{2}[/tex]? How does that work?
     
  5. Nov 2, 2008 #4

    gabbagabbahey

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    Yes, dividing both sides of the equation gives you:

    [tex]\frac{2x^2}{2t^2}=\frac{5t^2}{2t^2} \Rightarrow \frac{x^2}{t^2}=\frac{5}{2}[/tex]

    Then take the square root of both sides of the equation, that should give you x/t=...
     
  6. Nov 2, 2008 #5
    The answer is apparently [tex]\frac{5}{4}[/tex] though, and I don't understand where the 2t² comes from.
     
  7. Nov 2, 2008 #6

    statdad

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    Divide by [tex] 2 t^2 [/tex] to create an equation in which one side is constant, the other involving only [tex] x \text{ and } t [/tex].
     
  8. Nov 2, 2008 #7
    How do I get 5/4 then?
     
  9. Nov 2, 2008 #8

    statdad

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    I'm not sure, because looking back at your earlier posts (which I didn't do the first time) I think you have an early error.
    Here is your original equation.

    [tex]
    \sqrt{x^{2}-t^{2}}=2t-x
    [/tex]

    Notice that we cannot have [tex] x = 0 [/tex], since [tex] \sqrt{-t^2} [/tex] is not a real number. Since you want the value of [tex] x/t [/tex] we don't need to consider [tex] t = 0 [/tex] either.

    The first step in the solution is to square each side.

    [tex]
    x^2 - t^2 = \left(2t-x\right)^2 = 4t^2 - 4xt + x^2
    [/tex]

    The right hand side on your first step of the solution is only

    [tex]
    4t^2 - x^2
    [/tex]

    which isn't correct, since

    [tex]
    (2t-x)^2 \ne 4t^2 -x^2
    [/tex]

    Continuing on with the work,

    [tex]
    \begin{align*}
    x^2 - t^2 & = 4t^2 - 4xt + x^2 \\
    -t^2 & = 4t^2 - 4xt \\
    -5t^2 & = -4xt
    \end{align*}
    [/tex]

    Remembering you want to reach a point in which you have [tex] x/t [/tex] on one side,
    and a constant on the other, what do you need to do to the final equation here to reach it? (It is at this point that knowing [tex] t [/tex] cannot equal zero is important.)
     
  10. Nov 2, 2008 #9
    [tex]-5 = \frac{-4x}{t}[/tex]
    [tex]\frac{5}{4} = \frac{x}{t}[/tex]

    Oh ok I get it now, thanks so much.
     
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