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Math Help - Using Discriminant

  1. #1
    Member classicstrings's Avatar
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    Using Discriminant

    1. Show (k+1)x^2 - 2x - k = 0 has a solution for all values of k.

    So, I find discrim, = 4 + 4k^2 + 4k,

    Then I find the discrim of that = -48 = negative, meaning no solutions and graph of first discrim is above x-axis.

    How does prove that the original has solutions for all k?

    2. Show that ax^2 -(a+b)x + b = 0, has solution for all a, b.

    Discriminant is a^2 - 2ab + b^2, how do I interpret that?
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  2. #2
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    Quote Originally Posted by classicstrings View Post
    1. Show (k+1)x^2 - 2x - k = 0 has a solution for all values of k.

    So, I find discrim, = 4 + 4k^2 + 4k,
    Then I find the discrim of that = -48 = negative, meaning no solutions and graph of first discrim is above x-axis.
    How does prove that the original has solutions for all k?

    2. Show that ax^2 -(a+b)x + b = 0, has solution for all a, b.

    Discriminant is a^2 - 2ab + b^2, how do I interpret that?
    Hi, classicstring,

    to 1.) Could it be, that there are some restriction for k? Otherwise you have to set
    4 + 4k^2 + 4k ≥ 0 and solve for k.

    to 2.)

    a^2 - 2ab + b^2 = (a-b) ≥ 0

    EB
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by classicstrings View Post
    1. Show (k+1)x^2 - 2x - k = 0 has a solution for all values of k.

    So, I find discrim, = 4 + 4k^2 + 4k,
    Note that 4k^2 + 4k + 4 >= 0 for all k.

    -Dan
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  4. #4
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    Hello, classicstrings!

    1. Show (k+1)x - 2x - k .= .0 has a solution for all values of k.

    So, I find: .discrim. .= .4k + 4k + 4
    A quadratic has solutions if its discriminant is > 0.

    Can our discrimimant ever be negative?

    4k + 4k + 4 has a minimum point at its vertex: .k = -b/2a

    Hence, the vertex is at: .k .= .(-4)/(8) .= .-

    . . and the polynomial equals: .4(-) +4(-) + 4 .= .3

    The discriminant's minimum value is 3; hence, it is never negative.


    Therefore, the quadratic always has a solution.

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