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Thread: How is it possible to find the other root?

  1. #1
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    How is it possible to find the other root?

    In the cubic equation $\displaystyle x^3 + px^2 + qx + 26 = 0$ the constants p and q are real and $\displaystyle 2+3i$ is a root. I want to find the other two roots but I'm a little lost at what to do to get one of them.

    I know that $\displaystyle 2-3i$ is a root as the conjugate is always a root, and I also know that a cubic equation has either one or three real roots - so it follows that the other root I need to find is real.

    I tried finding the sum and product of the roots but I can't because I don't know what p and q are yet - I tried using an algebraic division method but the same can be said for this, and I don't know any other method to get the other root.

    The second part of the question asks to find the values of p and q, but I take it that there's a way to get the other root without getting these first? Could anyone suggest a method?

    Thanks if you can help me out
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  2. #2
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    Have you tried the sum and products of roots?

    Given roots $\displaystyle \alpha, \beta $ and $\displaystyle \gamma $ in the cubic equation $\displaystyle ax^3 + bx^2 + cx + d $

    $\displaystyle \alpha + \beta + \gamma = \frac{-b}{a} $

    $\displaystyle \alpha \beta + \alpha \gamma + \beta \gamma = \frac{c}{a} $

    $\displaystyle \alpha \beta \gamma = \frac{-d}{a} $
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  3. #3
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    Quote Originally Posted by db5vry View Post
    In the cubic equation $\displaystyle x^3 + px^2 + qx + 26 = 0$ the constants p and q are real and $\displaystyle 2+3i$ is a root. I want to find the other two roots but I'm a little lost at what to do to get one of them.

    I know that $\displaystyle 2-3i$ is a root as the conjugate is always a root, and I also know that a cubic equation has either one or three real roots - so it follows that the other root I need to find is real.
    Also, I presume you know that a cubic, with leading coefficient 1, is equal to $\displaystyle (x-\alpha)(x-\beta)(x- \gamma)$ where $\displaystyle \alpha$, $\displaystyle \beta$ and $\displaystyle \gamma$ are the roots. You know that one root is 2+ 3i and another is 2- 3i. That tells you that this can be written as $\displaystyle (x-2-3i)(x-2+3i)(x-\gamma)$ for some real number $\displaystyle \gamma$. Multiplying, $\displaystyle ((x-2)-3i)((x-2)+3i)= (x-2)^2-(3i)^2$$\displaystyle = (x-2)^2+ 9= x^2- 4x+ 4+ 9= x^2- 4x+ 13$. Now you know that $\displaystyle x^3+ px^2+ qx+ 26= (x^2- 4x+ 13)(x-\gamma)$. I would make a "wild guess" at $\displaystyle \gamma$ being 26/13= 2!.

    And once you know that $\displaystyle x^2+ px^2+ qx+ 26= (x^2- 4x+ 13)(x- 2)$ it is easy to find p and q.

    I tried finding the sum and product of the roots but I can't because I don't know what p and q are yet - I tried using an algebraic division method but the same can be said for this, and I don't know any other method to get the other root.

    The second part of the question asks to find the values of p and q, but I take it that there's a way to get the other root without getting these first? Could anyone suggest a method?

    Thanks if you can help me out
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  4. #4
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    Quote Originally Posted by Gusbob View Post
    Have you tried the sum and products of roots?

    Given roots $\displaystyle \alpha, \beta $ and $\displaystyle \gamma $ in the cubic equation $\displaystyle ax^3 + bx^2 + cx + d $

    $\displaystyle \alpha + \beta + \gamma = \frac{-b}{a} $ = -p

    $\displaystyle \alpha \beta + \alpha \gamma + \beta \gamma = \frac{c}{a} $ = q

    $\displaystyle \alpha \beta \gamma = \frac{-d}{a} $=-26
    I really don't see that going anywhere. Could you find a way to get the real root from the two that are already known?
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  5. #5
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    Quote Originally Posted by db5vry View Post
    In the cubic equation $\displaystyle x^3 + px^2 + qx + 26 = 0$ the constants p and q are real and $\displaystyle 2+3i$ is a root. I want to find the other two roots but I'm a little lost at what to do to get one of them.

    I know that $\displaystyle 2-3i$ is a root as the conjugate is always a root, and I also know that a cubic equation has either one or three real roots - so it follows that the other root I need to find is real.

    I tried finding the sum and product of the roots but I can't because I don't know what p and q are yet - I tried using an algebraic division method but the same can be said for this, and I don't know any other method to get the other root.

    The second part of the question asks to find the values of p and q, but I take it that there's a way to get the other root without getting these first? Could anyone suggest a method?

    Thanks if you can help me out
    You know that $\displaystyle 2 + 3i$ is a factor.

    So by the remainder and factor theorems, this means that

    $\displaystyle f(2 + 3i) = 0$

    So $\displaystyle (2 + 3i)^3 + p(2 + 3i)^2 + q(2 + 3i) + 26 = 0$

    $\displaystyle 8 + 36i - 54 - 27i + p(4 + 12i - 9) + 2q + 3qi + 26 = 0$

    $\displaystyle -20 + 9i - 5p + 12pi + 2q + 3qi = 0$

    $\displaystyle 2q - 5p + i(12p + 3q) = 20 + 9i$

    Equating real and imaginary components gives

    $\displaystyle 2q - 5p = 20$ and $\displaystyle 12p + 3q = 9$.

    Solve these equations simultaneously.
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  6. #6
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    Quote Originally Posted by db5vry View Post
    In the cubic equation $\displaystyle x^3 + px^2 + qx + 26 = 0$ the constants p and q are real and $\displaystyle 2+3i$ is a root. I want to find the other two roots but I'm a little lost at what to do to get one of them.

    I know that $\displaystyle 2-3i$ is a root as the conjugate is always a root, and I also know that a cubic equation has either one or three real roots - so it follows that the other root I need to find is real.

    I tried finding the sum and product of the roots but I can't because I don't know what p and q are yet - I tried using an algebraic division method but the same can be said for this, and I don't know any other method to get the other root.

    The second part of the question asks to find the values of p and q, but I take it that there's a way to get the other root without getting these first? Could anyone suggest a method?

    Thanks if you can help me out
    To find the third root, you know that

    $\displaystyle x - (2 + 3i)$ and $\displaystyle x - (2 - 3i)$ are factors of the cubic.

    Therefore $\displaystyle [x - (2 + 3i)][x - (2 - 3i)]$ is also a factor.

    $\displaystyle =(x - 2 - 3i)(x - 2 + 3i)$

    $\displaystyle = x(x - 2 + 3i) - 2(x - 2 + 3i) - 3i(x - 2 + 3i)$

    $\displaystyle = x^2 - 2x + 3xi - 2x + 4 - 6i - 3xi + 6i + 9$

    $\displaystyle = x^2 - 4x + 13$


    So now divide your cubic (and from my previous post you will have found $\displaystyle p$ and $\displaystyle q$) by $\displaystyle x^2 - 4x + 13$.

    Then you will have the other factor, and solving this $\displaystyle = 0$ for $\displaystyle x$ will give you the third root.
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    Quote Originally Posted by db5vry View Post
    I really don't see that going anywhere. Could you find a way to get the real root from the two that are already known?
    Post #3 told you exactly what to do and then did it for you.
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