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Math Help - Can you help me solve this equation?

  1. #1
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    Can you help me solve this equation?

    1. Can the formula be solved for either r or t? Explain. 2. if the formula can be solved for r, what does r=? 2-3rt+t=3t^2-t+r^2t^2-5r^2
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  2. #2
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    Yes, it can be solved for both r and t.

    Do you know how to rearrange the equation into the form of ax^2 + bx + c = 0 and put it onto the quadratic formula?
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  3. #3
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    Usually I am able to but this one just baffles me
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  4. #4
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    Here's a starter for you: (this one is for the variable t)

    2-3rt+t=3t^2-t+r^2t^2-5r^2

    2-3rt+t-3t^2+t-r^2t^2+5r^2=0

    -3t^2-r^2t^2-3rt+2t+5r^2+2=0

    t^2(-3-r^2)+t(-3r+2)+5r^2+2=0

    Now can you solve for the variable t?
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  5. #5
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    is the answer t=3r+2 +or- the square root of (-3r+2)^2-4(-3-r^2)(5r^2+2) over 2(-3-r^2) ?
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  6. #6
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    Yes. That is correct.

    Now can you solve for yourself for the variable r?
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  7. #7
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    would r=3t^2 +or- the square root of (-3t+2t)^2-4(-3t^2)(5r^2+2) over 2(-3t^2) ?
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  8. #8
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    No. It would help if you showed some working as well so we can see where you went wrong.
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  9. #9
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    I took -3t^2-r^2t^2-3rt+2t+5r^2+2=0 and converted it to r^2(-3t^2)+r(-3t+2t)+5r^2+2=0 and then turned that into r=3t^2 +or- the square root of (-3t+2t)^2-4(-3t^2)(5r^2+2) over 2(-3t^2)
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  10. #10
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    Quote Originally Posted by MizzCyanide View Post
    I took -3t^2-r^2t^2-3rt+2t+5r^2+2=0 and converted it to r^2(-3t^2)+r(-3t+2t)+5r^2+2=0
    I think you factorised wrong... maybe you should try again? As I can see from the equation, there are two r^2 parts. So where'd the second part go?

    Also it may pay to learn LaTeX so it is easier to understand.
    LaTeX Tutorial
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  11. #11
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    ok lets try with LaTeX

    I took -3t^2-r^2t^2-3rt+2t+5r^2+2=0
    r^2(-3t^2)+r(-3t+2t)+5r^2+2=0
    r=3t^2 +-\sqrt(-3t+2t)^2-4(-3t^2)(5r^2+2)/2(-3t^2)
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  12. #12
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    You factorised wrong. Can you see where you went wrong?
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  13. #13
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    not really :/
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  14. #14
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    r^2(-3t^2)+r(-3t+2t)+5r^2+2=0

    Now can you see that 5r^2 has an r squared variable in it? It should be factorised along with the other r squared variable.
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  15. #15
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    oh...so it should instead be  r^2(5+2) so then the (5+2)=7? so in that case would the (-3t^2) and (5+2)=7 be combined into (-3t^2+7)? or would they just be separated to create r=3t^2+-\sqrt(-3t+2t)^2-4(-3t^2)(7)/2(-3t^2)
    Last edited by MizzCyanide; October 6th 2010 at 09:17 PM.
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