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Thread: 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 $\displaystyle y=x^3$, then we have:

    $\displaystyle y^2-ky+9=0$

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

    $\displaystyle
    x=\pm \sqrt{\left(
    \frac{k\pm\sqrt{k^2-36}}{2}
    \right)}
    $

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

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

    so:

    $\displaystyle
    3\ \sqrt{\left(
    \frac{k -\sqrt{k^2-36}}{2}
    \right)}=\sqrt{\left(
    \frac{k +\sqrt{k^2-36}}{2}
    \right)}
    $

    Alternative method:

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

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

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

    .
    Last edited by Constatine11; Jan 2nd 2009 at 09:53 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 $\displaystyle a,\ ra,\ r^2a$ for some real $\displaystyle a$ and $\displaystyle r$.

    So:

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

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

    .
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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
    MHF Contributor red_dog's Avatar
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    a) Let $\displaystyle x_1=a-3r, \ x_2=a-r, \ x_3=a+r, \ x_4=a+3r$
    Then $\displaystyle x_1+x_2+x_3+x_4=0\Rightarrow 4a=0\Rightarrow a=0$
    $\displaystyle x_1x_2x_3x_4=9\Rightarrow 9a^4=9\Rightarrow a=\pm 1$

    Now find k
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  7. #7
    MHF Contributor red_dog's Avatar
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    b) Let $\displaystyle x_1=\frac{a}{q}, \ x_2=a, \ x_3=aq$
    Then $\displaystyle x_1x_2x_3=-64\Rightarrow a^3=-64\Rightarrow a=-4$
    $\displaystyle x_1+x_2+x_3=6\Rightarrow -\frac{4}{q}-4-4q=6$
    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 $\displaystyle k$ algebraically.

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

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


    From Vieta's formulas . . .

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

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



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

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

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


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



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

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

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  9. #9
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    Thank you so much!
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