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Quadratics homework help

  1. Jul 14, 2011 #1
    1. The problem statement, all variables and given/known data

    I am meant to look for both points when thiobject is 6m above the ground

    the equation is

    -5x^2+10x+2=6

    my question is how do i solve it for specific points in time,
    Do i just make the y term 0 and solve from there?



    2. Relevant equations



    3. The attempt at a solution

    -5x^2 +10x -4

    -5(x-1)^2-5=0
     
  2. jcsd
  3. Jul 14, 2011 #2
    Re: Quadratics!

    I'm guessing [itex] y [/itex] is the vertical displacement and [itex] x [/itex] is some other parameter (the point or points you want to find at which [itex] y = 6[/itex]m). Thus you need to solve for [itex] x [/itex] given [itex] y = 6 [/itex]m.

    So all you need to do is solve the equation for [itex] x [/itex] like you have already started doing (either using the quadratic equation or by factorizing like you have done). You have however made a slight mistake in your factoring.

    It should be

    [itex] -5{(x-1)}^2 + 1=0 [/itex]

    Multiply this out and you should see that this gives you the original expression.

    From this it is clear that (subtract 1 and divide by -5 on both sides).

    [itex] {(x-1)}^2 = \frac{1}{5} [/itex]

    [itex] x-1 = \pm \sqrt{\frac{1}{5}} [/itex]

    [itex] x = 1 \pm \sqrt{\frac{1}{5}} [/itex]

    You could also have used the quadratic equation:

    [itex] x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} [/itex]

    To solve the general quadratic: [itex] ax^2 + bx + c = 0 [/itex].
    Thus in your case we solve for x like so...

    [itex] x = \frac{-10 \pm \sqrt{10^2-4(-5)(-4)}}{2(-5)} [/itex]

    [itex] = \frac{-10 \pm \sqrt{100-80}}{-10} [/itex]

    [itex] = \frac{-10 \pm \sqrt{20}}{-10} [/itex]

    [itex] = 1 \mp \sqrt{\frac{20}{100}} [/itex]

    [itex] = 1 \mp \sqrt{\frac{1}{5}} [/itex]
     
    Last edited: Jul 14, 2011
  4. Jul 14, 2011 #3
    Re: Quadratics!

    Thank you for the help, I am just curious as to where you get the +1 constant from.

    nvm, after revision i found your answer to be correct.
    You have my gratitude. :D
     
    Last edited: Jul 14, 2011
  5. Jul 14, 2011 #4
    Re: Quadratics!

    In case the +1 is still bothering you, consider first the expression that you ended up with.

    [itex] -5{(x-1)}^2-5=0 [/itex]

    This expression should be consistent with the original, when one multiplies out to see if it does agree with the initial equation one gets.

    [itex] -5x^2 -5(-2x) -5(1) -5= 0 [/itex]

    [itex] -5x^2 +10x -5 -5= 0 [/itex]

    We see now that in order to get -4 and not -10 we need to replace the -5 with a +1.
     
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