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Math Help - CSET math question

  1. #1
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    Oct 2008
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    Question CSET math question

    Find b such that f(x) and g(x) intersect at one point only.
    f(x) = 3x^2 + 5x + 7
    g(x) = bx + 3

    ____________________________________________

    I set them equal
    3x^2 + 5x + 7 = bx + 3
    3x^2 + (5-b)x + 4 = 0

    -(5-b) + - sqrt b^2 - 10b + 25 - (48)
    -------------------------------------
    6

    -5 + b + - sqrt b2 -10b -23 / 6 I don't know what to do next the answer is suppose to be 5 + - 4 sqrt 3

    How did they get rid of the "b" in the sqrt

    I would love some help
    Thanks,
    Marion
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  2. #2
    Junior Member
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    A polynomial and a linear function intersect only once if the linear function is a tangent line.
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  3. #3
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    Oct 2008
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    I need more help

    I am sorry to bug but I still don't understand what I need to do next to solve the problem and if I am on the right track.
    Please give additional help.
    Marion
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  4. #4
    Junior Member
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    You have to find a tangent line to f(x)=3x^2 + 5x + 7 such that it has the form bx+3.
    The slope of the tangent line is given by f'(x)=6x+5, so that at a point y the slope is 6y+5 and the tangent line is given by h(x)=(6y+5)x+c.
    Also at that point y the value of the tangent is equal to the value of the polynomial in that point so that you have h(y)=f(y). Now you look for the values of y such that c=3, then you have h(y)=(6y+5)y+3 and f(y)=3y^2+5y+7. These should be equal, so you find y=+/- sqrt(4/3).
    So b=6(+/- sqrt(4/3))+5 = 5+/-sqrt(48) = 5+/-4sqrt(3).
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