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Math Help - Exercise - polynomial equation and geometric sequence

  1. #1
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    Exercise - polynomial equation and geometric sequence

    Hey, my neighbour again send me one exercise.

    1. In the equation x^{3} - 7x^{2} - 21x + a = 0 you must find solutions whose are in geometric sequence. For which a?

    How can I find this a?

    2. In the equation x^{4} -(a+3)x^2 + (a + 2) = 0 must find this a that solutions are in arithmetic sequence.

    How can I find this a?
    Last edited by lebdim; August 12th 2014 at 11:51 AM.
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  2. #2
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    Re: Exercise - polynomial equation and geometric sequence

    For the first question, you actually have to look up the formula for finding cubic roots.
    Here's the thing: let the three roots be χ1, χ2 and χ3, then it's required that χ1:χ2=χ2:χ3. Common ratio right lol?
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  3. #3
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    Re: Exercise - polynomial equation and geometric sequence

    For the second question, Δ=(a+3)^2-4(a+2)=a^2+6a+9-4a-8=a^2+2a+1=(a+1)^2
    First, x^2=(1/2)[(a+3)+/-(a+1)] = (1/2)[(2a+4) or 2]= a+2 or 1.
    From x^2= 1 we know x=-1 or 1.
    Then there are 2 possibilities:
    (1) 0<a+2<1 then -1< -sqrt(a+2) < sqrt(a+2) < 1, in which case, you have -sqrt(a+2)-(-1)=sqrt(a+2)-[-sqrt(a+2)].
    (2) a+2>1 then -sqrt(a+2) < -1 < 1 < sqrt(a+2).
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  4. #4
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    Re: Exercise - polynomial equation and geometric sequence

    No, we don't need the general formula, here. Suppose our roots are $r_1, r_2,r_3$ in geometric sequence.

    Then $r_2 = cr_1$, and $r_3 = cr_2 = c^2r_1$. Let us just use $r$ for $r_1$. So our polynomial factors as:

    $x^3 - 7x - 21x + a = (x - r)(x - cr)(x - c^2r) = x^3 - (r + cr + c^2r)x^2 + (cr^2 + c^2r^2 + c^3r^2)x - c^3r^3$.

    Evidently, then:

    $r(1 + c + c^2) = 7$
    $cr^2(1 + c + c^2) = -21$
    $a = -c^3r^3$

    From this we see that $cr = -3$. This is all we need to find $a$, since $a = -c^3r^3 = -(cr)^3 = -(-27) = 27$.

    But we can do more:

    Since $7 = r(1+c+c^2) = \dfrac{-3}{c}(1+c+c^2) = -\dfrac{3}{c} - 3 -3c$,

    multiplying by $c$, we get:

    $-3c - 3 - 3c^2 = 7 \implies c^2 + \dfrac{10}{3}c + 1 = 0$.

    From the quadratic formula, we obtain:

    $c = \dfrac{-\frac{10}{3} \pm \sqrt{\frac{100}{9} - 4}}{2} = -\dfrac{10}{6} \pm \dfrac{\sqrt{\frac{64}{9}}}{2} = -\dfrac{5}{3} \pm \dfrac{4}{3}$

    so $c = -3$ or $c = -\dfrac{1}{3}$.

    If $c = -3$ then since $cr = -3$ we must have $r = 1$. This works, because $1(1 - 3 + 9) = 1(7) = 7$. Thus our roots are $1,-3,9$.

    If $c = -\dfrac{1}{3}$ then $r = 9$. This means our roots are $9,-3,1$, which are the same set of roots.

    It is easy to see that:

    $(x - 1)(x + 3)(x - 9) = x^3 - 7x^2 - 21x + 27$, as required.
    Thanks from dennydengler
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  5. #5
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    Re: Exercise - polynomial equation and geometric sequence

    Here's a solution to your first question:

    Last edited by johng; August 12th 2014 at 06:15 PM.
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