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  1. #1
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    a question

    for what values of the parameter m there will be
    4 answers in wich can be arranged so the subtraction of each 2 answers
    will be permanent
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  2. #2
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    Quote Originally Posted by fima2001 View Post
    for what values of the parameter m there will be
    4 answers in wich can be arranged so the subtraction of each 2 answers
    will be permanent
    Maybe someone else will be able to tell us what this means, but to me
    it seems incomprehensible. Can you clarify what it is you are asking?

    Thanks

    RonL
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by fima2001 View Post
    for what values of the parameter m there will be
    4 answers in wich can be arranged so the subtraction of each 2 answers
    will be permanent
    To start you off, this is a biquadratic equation. So:

    x^4 + 6mx^2 + (3m+8)^2 = 0

    You can solve this for x^2 using the quadratic formula:

    x^2 = \frac{-6m \pm \sqrt{36m^2 - 4(3m+8)^2}}{2}

    x^2 = \frac{-6m \pm \sqrt{36m^2 - 36m^2 - 192m - 256}}{2}

    x^2 = \frac{-6m \pm \sqrt{-192m - 256}}{2}

    x^2 = \frac{-6m \pm 16 \sqrt{-3m - 4}}{2}

    x^2 = -3m \pm 8 \sqrt{-3m - 4}

    Now take the square root of both sides to get your 4 solutions:
    x = \pm \sqrt{ -3m \pm 8 \sqrt{-3m - 4}} (Where the two \pm symbols are independant of each other.)

    To continue we need to know what you mean by "the subtraction of each 2 answers will be permanent."

    -Dan
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  4. #4
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    Hello, all!

    I think I have an interpretation . . . and a solution.

    How about this?


    For what values of the parameter m will the equation
    have four roots in arithmetic progression?

    . . x^4 + 6mx^2 + (3m+8)^2\;=\;0

    Let the four roots be: . a,\:a+d,\:a+2d,\:a+3d

    The sum of the roots is equal to the negative of the coefficient of x^3.
    . . Hence: .  a + (a+d)+(a+2d)+(a+3d) \:=\:0
    . . And we have: . 4a + 6d\:=\:0\quad\Rightarrow\quad d = -\frac{2}{3}a [1]

    The product of the roots is equal to the constant term.
    . . Hence: . a(a+d)(a+2d)(a+3d) \:=\:(3m+8)^2

    Substitute [1]: . a\left(a - \frac{2}{3}a\right)\left(a - \frac{2}{3}a\right)\left(a - 2a\right)\;=\;(3m+8)^2

    . . a\left(\frac{a}{3}\right)\left(-\frac{a}{3}\right)\left(-a\right) \:=\:(3m+8)^2\quad\Rightarrow\quad a^4 = 9(3m+8)^2\quad\Rightarrow\quad a^2 = 3(3m+8)^2 [2]


    The sum of the the roots, taken in pairs, is the coefficient of x^2.
    a(a+d)+a(a+2d)+a(a+3d)+(a+d)(a+2d)+ (a+d)(a+3d)+(a+2d)(a+3d)\:=\:6m

    Substitute [1]: . a\left(\frac{a}{3}\right) + a\left(-\frac{a}{3}\right) + a(-a) + \left(\frac{a}{3}\right)\left(-\frac{a}{3}\right) + \left(\frac{a}{3}\right)\left(-a\right) + \left(-\frac{a}{3}\right)\left(-a\right) \:=\:6m

    and we have: . -\frac{10}{9}a^2\:=\:6m\quad\Rightarrow\quad a^2 \:=\:-\frac{27}{5}m [3]


    Equate [2] and [3]: . 3(3m + 8) \:=\:-\frac{27}{5}m\quad\Rightarrow\quad \boxed{m = -\frac{5}{3}}

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    Check

    From [3]: . a^2\:=\:-\frac{27}{5}\left(-\frac{5}{3}\right)\:=\;9\quad\Rightarrow\quad a = \pm3

    And from [1]: . d \:=\:-\frac{2}{3}(\pm3)\:=\:\mp2

    Therefore, the roots are: . -3,\:-1,\:1,\:3

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  5. #5
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    i ment

    there will be 4 answers and the subtraction of each 2 close
    roots will be the same resolt
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