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    a question about quadratic equetion

    for what values of M

    i added the question in a picture
    a question about quadratic equetion-question.gif
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  2. #2
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    Quote Originally Posted by fima2001 View Post
    for what values of M

    i added the question in a picture
    Click image for larger version. 

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    I assume you mean to solve for when y = 0 and x1, x2 are the two roots?

    0 = (m-2)x^2 - 2(m-2)x + (4-m)

    So
    a = m-2
    b = -2(m-2) = -2m + 4
    c = 4 - m

    x = [-b (+/-) sqrt(b^2 - 4*a*c)]/(2a)

    x = [-(-2m+4) (+/-) sqrt({-2m+4}^2 - 4*(m-2)*(4-m))]/(2*(m-2))

    x = [2m-4 (+/-) sqrt(4m^2 - 16m + 16 - 4*(-m^2 +6m - 8))]/(2m-4)

    x = [(2m-4) (+/-) sqrt(8m^2 - 40m + 48)]/(2m+4)

    x = [(2m-4) (+/-) 2*sqrt(2m^2 - 10m + 12)]/(2m+4)

    x = 1 (+/-) [2/(2m+4)]*sqrt(2m^2 - 10m + 12)

    x = 1 (+/-) sqrt(2m^2 - 10m + 12)/(m+2)

    So the larger of these, x2, must use the "+" and x1 must use the "-".

    Consider the x2 condition:
    2<x2<6

    Thus
    2 < 1 + sqrt(2m^2 - 10m + 12)/(m+2) < 6

    or

    1 < sqrt(2m^2 - 10m + 12)/(m+2) < 5
    This has critical points where the numerator and denominator are zero. Thus it has critical points at m = -2, m = 2, and m = 3. So split the number line into 4 intervals and check the inequality:
    (-infinity, -2): No.
    (-2, 2): Yes
    (2, 3): No.
    (3, infinity): No.

    So we know that -2 < m < 2.

    Now let's look at x1:
    1 < x1 < 2

    1 < 1 - sqrt(2m^2 - 10m + 12)/(m+2) < 2

    0 < - sqrt(2m^2 - 10m + 12)/(m+2) < 1

    Again we find the critical points at m = -2, 2, and 3. We already know that -2 < m < 2, so all we need to do is check the inequality on this interval. Also note that m must be negative in order for this to hold. So I'm only going to check the inequality on (-2, 0).

    (-2, 0): No.

    Thus it is my conclusion that there is no such m giving roots x1 and x2 the desired properties. (I have verified this for several representative values of m.)

    -Dan
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