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Homework Help: When does this derivative equal 7

  1. Oct 4, 2009 #1
    1. The problem statement, all variables and given/known data

    Solving a derivative for when it equals to 7.

    2. Relevant equations

    [tex] p'(t) = {88t+210}/{2\sqrt{44t^{2}+210t}[/tex]

    3. The attempt at a solution

    [tex]7(2\sqrt{44t^{2}+210t}) = {88t+210}[/tex]

    This is where I get stuck. Do I divied both sides by 7 then again by 2? then I would square both sides to get (88/14 - 15)^{2}. Then I expand, set it to 0 and use quadratic formula?
  2. jcsd
  3. Oct 4, 2009 #2


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    Homework Helper

    Re: Quadratic

    I'm a bit confused.
    At first glance I would assume this means "find the derivative where the variable t=7" but from what you've done, I guess it must instead mean "find the value(s) of t where the derivative is 7". Is this correct?

    Now, without using parenthesis it's impossible to be certain what the derivative function actually is:
    but judging by your last line:
    I'm now assuming you mean [tex]p'(t) = \frac{88t+210}{2\sqrt{44t^{2}+210t}}[/tex]
    Again, is this correct?

    Finally, if all is correct thus far, yes that is exactly how you would go about solving it :smile: But remember to check your answers because once you square both sides, you might have extra solutions that don't work.
  4. Oct 4, 2009 #3
    Re: Quadratic

    Yes! that is what i meant :).

    The problem is, though, when I get to this:

    [tex]7(2\sqrt{44t^{2}+210t}) = {88t+210}[/tex]

    I divide out the seven on both side to get
    [tex]2\sqrt{44t^{2}+210t} = {88/7t+30}[/tex]

    then again by 2

    [tex]\sqrt{44t^{2}+210t} = {88/14t+15}[/tex]

    Now, when I square both side? do I write it like this:[tex]{44t^{2}+210t = ({88/14t+15})^{2}[/tex]

    which I just expand out?
    Last edited: Oct 4, 2009
  5. Oct 4, 2009 #4
    Re: Quadratic

    You can expand it. Don't forget that you can reduce 88/14.
  6. Oct 4, 2009 #5
    Re: Quadratic

    How come I can't just take the square of 44t/14 and 15?
  7. Oct 4, 2009 #6
    Re: Quadratic

    because (a+b)^2 does not equal a^2+b^2
    in fact it is: a^2+2ab+b^2
    try with numbers. (3+4)^2 doesnt equal 3^2+4^2
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