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Math Help - Solving for roots and coefficient

  1. #1
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    Solving for roots and coefficient

    Solve for the roots and the coefficient k algebraically.
    a)The roots of f(x)=x^4 - kx^2 + 9=0 are in arithmetic progression.

    And a second problem (same kind)
    b) The roots of x^3 - 6x^2 + kx + 64=0 are in geometric progression.

    Thanks for the help. ^^
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  2. #2
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    Quote Originally Posted by Jukixel View Post
    Solve for the roots and the coefficient k algebraically.
    a)The roots of f(x)=x^4 - kx^2 + 9=0 are in arithmetic progression.
    This can have 0, 2 or 4 real roots, for them to be arithmetic progression there must be 4.

    put y=x^3, then we have:

    y^2-ky+9=0

    and for the original equation to have four real roots this must have two real roots. These are:

     <br />
x=\pm \sqrt{\left(<br />
\frac{k\pm\sqrt{k^2-36}}{2}<br />
\right)}<br />

    for these to be real and distinct we require that k>6.

    For these to be in arithmetic progression there must be an a such that these roots are -3a,-a,a,3a

    so:

     <br />
3\ \sqrt{\left(<br />
\frac{k -\sqrt{k^2-36}}{2}<br />
\right)}=\sqrt{\left(<br />
\frac{k +\sqrt{k^2-36}}{2}<br />
\right)}<br />

    Alternative method:

    The roots must be a,\ a+b,\ a+2b,\ a+3b, for some real a and b so:

    x^4 - kx^2 + 9=(x-a)(x-(a+b))(x-(a+2b))(x-(a+3b))

    Now expand the product on the right, and equate the coefficients of like powers on the left and right to solve for k,\ a,\ b

    .
    Last edited by Constatine11; January 2nd 2009 at 10:53 AM.
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  3. #3
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    Sorry, I don't really get the last step. =/
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  4. #4
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    Quote Originally Posted by Jukixel View Post
    And a second problem (same kind)
    b) The roots of x^3 - 6x^2 + kx + 64=0 are in geometric progression.

    Thanks for the help. ^^
    Here the roots are of the form a,\ ra,\ r^2a for some real a and r.

    So:

    x^3 - 6x^2 + kx + 64=(x-a)(x-ra)(x-r^2a)

    Now expand the product on the right and equate coefficients of like powers of x to solve for k,\ a,\ r

    .
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  5. #5
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    Oh, I think I get this method. ^^ Thanks a lot!
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  6. #6
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    a) Let x_1=a-3r, \ x_2=a-r, \ x_3=a+r, \ x_4=a+3r
    Then x_1+x_2+x_3+x_4=0\Rightarrow 4a=0\Rightarrow a=0
    x_1x_2x_3x_4=9\Rightarrow 9a^4=9\Rightarrow a=\pm 1

    Now find k
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  7. #7
    MHF Contributor red_dog's Avatar
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    b) Let x_1=\frac{a}{q}, \ x_2=a, \ x_3=aq
    Then x_1x_2x_3=-64\Rightarrow a^3=-64\Rightarrow a=-4
    x_1+x_2+x_3=6\Rightarrow -\frac{4}{q}-4-4q=6
    Now solve for q, then find k.
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  8. #8
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    Hello, Jukixel!

    Solve for the roots and the coefficient k algebraically.

    a) The roots of f(x)\:=\:x^4 - kx^2 + 9\:=\:0 are in arithmetic progression.
    The roots are: . a,\:a+d,\:a+2d,\:a+3d

    The function is: . f(x) \:=\:x^4 + 0\!\cdot\!x^3 - kx^2 + 0\!\cdot\!x + 9 \:=\:0


    From Vieta's formulas . . .

    . . a + (a+d) + (a+2d)+(a+3d) \:=\:0 \quad\Rightarrow\quad d \:=\:\text{-}\tfrac{2}{3}a .[1]

    . . a(a+d)(a+2d)(a+3d) \:=\:9 .[2]



    Substitue [1] into [2]: . a\left(a-\tfrac{2}{3}a\right)\left(a - \tfrac{4}{3}a\right)(a-2a) \:=\:9

    . . a\left(\tfrac{1}{3}a\right)\left(\text{-}\tfrac{1}{3}a\right)\left(\text{-}a\right) \:=\:9 \quad\Rightarrow\quad \tfrac{1}{9}a^4 \:=\:9\quad\Rightarrow\quad a^4 \:=\:81 \quad\Rightarrow\quad a \:=\:\pm3

    Substitute into [1]: . d \:=\:-\tfrac{2}{3}\left(\pm3\right) \:=\:\mp2


    . . Therefore, the roots are: . \boxed{\text{-}3,\:\text{-}1,\:1,\:3}



    And the polynomial is: . f(x) \:=\:(x+3)(x+1)(x-1)(x-3) \:=\:x^4 - 10x^2 + 9

    . . Therefore: . \boxed{k \:=\:10}

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  9. #9
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    Thank you so much!
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