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Math Help - intersection of y=mx+k and y=ax^2+bx+c where a=/=0

  1. #1
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    intersection of y=mx+k and y=ax^2+bx+c where a=/=0

    determine intersection of y=mx+k and y=ax^2+bx+c where a=/=0

    i got

    ((-b+m+sqrt(-4a(c-k)+(b-m)^2))/2a, (m(-b+m+sqrt(-4a(c-k)+(b-m)^2)+k)/2a))

    and

    ((-b+m-sqrt(-4a(c-k)+(b-m)^2))/2a, (m(-b+m-sqrt(-4a(c-k)+(b-m)^2)+k)/2a))

    as my 2 solutions

    is it correct?
    Last edited by iragequit; March 8th 2011 at 09:08 PM.
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  2. #2
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    Its hard to decodee your solution but it looks like your on the right track.

    For \displaystlye mx+k= ax^2+bx+c \implies 0=ax^2+(b-m)x+c-k

    Therefore \displaystlye x= \frac{-(b-m)^2\pm\sqrt{(b-m)^2-4a(c-k)}}{2a}
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  3. #3
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    Quote Originally Posted by pickslides View Post
    Its hard to decodee your solution but it looks like your on the right track.

    For \displaystlye mx+k= ax^2+bx+c \implies 0=ax^2+(b-m)x+c-k

    Therefore \displaystlye x= \frac{-(b-m)^2\pm\sqrt{(b-m)^2-4a(c-k)}}{2a}
    why woudl you square -(b-m)? i'm gusesing that's a mistake right?

    also, shoudl i leave it at -(b-m) or simplify / expand to -b+m +- sqrt(...etc..etc) ?

    also, what's y equal to?
    i subbed in x into y=mx+k and got y= m(-b+-sqrt(-4a(c-k)+(b-m)^2)/2a+k

    i don't know how to simplify it after you sub it in

    also it would be helpful if you could answer the rest of the questions in first post, thanks
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  4. #4
    Forum Admin topsquark's Avatar
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    Yes the -(b - m)^2 is a typo.

    Take a look at the discriminant. Is this always non-negative? This is your extra criterion for the existence of a solution.

    Just a thought. I wouldn't worry too much about an explicit form for y. Calculate and simplify x then feed that value of x into y = mx + k. Algebraically it will be simpler to deal with (as well as easier to write your solution.)

    -Dan
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  5. #5
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    nbm
    Last edited by iragequit; March 8th 2011 at 03:57 PM.
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