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Math Help - an ambiguous quadratic equation question.

  1. #1
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    an ambiguous quadratic equation question.

    This question is very unclear as to what it wants for an answer, so I'd just like to check and see if my working is satisfactory.

    The question:
    Solve the equation x^3+2x^2+x+2=0 if -2 is a root.

    All I did with my working is replace x with -2.
    Is this going to satisfy the question?
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  2. #2
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    Quote Originally Posted by quikwerk View Post
    This question is very unclear as to what it wants for an answer, so I'd just like to check and see if my working is satisfactory.

    The question:
    Solve the equation x^3+2x^2+x+2=0 if -2 is a root.

    All I did with my working is replace x with -2.
    Is this going to satisfy the question?
    Seems like a safe bet to me.

    Also, this is a cubic not a quadratic.

    If you are feeling ambitious, you can use synthetic division to take the equation down to a quadratic and find the remaining two roots.
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  3. #3
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    Quote Originally Posted by dwsmith View Post
    Seems like a safe bet to me.

    Also, this is a cubic not a quadratic.

    If you are feeling ambitious, you can use synthetic division to take the equation down to a quadratic and find the remaining two roots.
    I wouldn't even do that. Just equate coefficients.

    If x=-2 is a root, then (x+2) is a factor.

    x^3+2x^2+x+2=(x+2)(x^2 + ...)
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  4. #4
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    Quote Originally Posted by dwsmith View Post
    Seems like a safe bet to me.

    Also, this is a cubic not a quadratic.

    If you are feeling ambitious, you can use synthetic division to take the equation down to a quadratic and find the remaining two roots.
    I'll probably find the other roots to be safe, but if I were to just go with my original working, all I would have to do is establish that the equation is true with -2 replacing x, yes?
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  5. #5
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    Quote Originally Posted by quikwerk View Post
    I'll probably find the other roots to be safe, but if I were to just go with my original working, all I would have to do is establish that the equation is true with -2 replacing x, yes?
    Yes, just show it is true when x = -2.
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  6. #6
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    Quote Originally Posted by quikwerk View Post
    I'll probably find the other roots to be safe, but if I were to just go with my original working, all I would have to do is establish that the equation is true with -2 replacing x, yes?
    No!!!

    "Solve" the equation means

    find all x for which the equation is true

    hence, you must use the clue given to find the remaining solutions.

    "Solve" means "find the solutions".

    One solution is x=-2

    (x+2)\left(x^2+bx+c\right)=x^3+2x^2+x+2

    Clearly, this is zero if x=-2 since these are two ways to write the same expression.

    If we multiply this out, 2c=2\Rightarrow\ c=1

    bx^2+2x^2=2x^2\Rightarrow\ b=0

    (x+2)\left(x^2+1\right)=x^3+x+2x^2+2


    The other roots are complex.
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  7. #7
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    Quote Originally Posted by Archie Meade View Post
    No!!!

    "Solve" the equation means

    find all x for which the equation is true

    hence, you must use the clue given to find the remaining solutions.

    "Solve" means "find the solutions".

    One solution is x=-2

    (x+2)\left(x^2+bx+c\right)=x^3+2x^2+x+2

    Clearly, this is zero if x=-2 since these are two ways to write the same expression.

    If we multiply this out, 2c=2\Rightarrow\ c=1

    bx^2+2x^2=2x^2\Rightarrow\ b=0

    (x+2)\left(x^2+1\right)=x^3+x+2x^2+2


    The other roots are complex.
    My hopes are dashed!

    Ok, so now I have to solve this cubic equation...
    Would I be correct in supposing that one of the roots might be 2?
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  8. #8
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    Quote Originally Posted by quikwerk View Post
    My hopes are dashed!

    Ok, so now I have to solve this cubic equation...
    Would I be correct in supposing that one of the roots might be 2?
    No, one of the roots is x=-2

    and it's the only real root.

    There are also two complex roots, which are the solution to the 2nd factor being zero.

    x^3+2x^2+x+2=(x+2)\left(x^2+1\right)

    x^3+2x^2+x+2=0\Rightarrow\ (x+2)\left(x^2+1\right)=0\Rightarrow\ x+2=0\;\;\;or\;\;\;x^2+1=0

    (x+2)=0\Rightarrow\ x=-2

    If you are familiar with complex numbers, then you need to solve x^2+1=0

    Otherwise, you need to be aware that x=-2 is the only real root

    since there is no real solution to x^2+1=0

    since x^2=-1 is the solution to that.
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  9. #9
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    Quote Originally Posted by Archie Meade View Post
    No, one of the roots is x=-2

    and it's the only real root.

    There are also two complex roots, which are the solution to the 2nd factor being zero.

    x^3+2x^2+x+2=(x+2)\left(x^2+1\right)

    x^3+2x^2+x+2=0\Rightarrow\ (x+2)\left(x^2+1\right)=0\Rightarrow\ x+2=0\;\;\;or\;\;\;x^2+1=0

    (x+2)=0\Rightarrow\ x=-2

    If you are familiar with complex numbers, then you need to solve x^2+1=0

    Otherwise, you need to be aware that x=-2 is the only real root

    since there is no real solution to x^2+1=0

    since x^2=-1 is the solution to that.
    Ok, I understand that, so what you are saying is that one root is "i" and another is something multiplied by i?
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  10. #10
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    Yes, and that "something" can only be one value.

    What is (-i)^2\;\;?

    Or

    (x+i)(x-i)=\;\;?
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  11. #11
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    Quote Originally Posted by Archie Meade View Post
    Yes, and that "something" can only be one value.

    What is (-i)^2\;\;?

    Or

    (x+i)(x-i)=\;\;?
    i*i=-1
    So you are saying that the solution set is:{-2, i, i^2}?
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  12. #12
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    Quote Originally Posted by quikwerk View Post
    i*i=-1
    So you are saying that the solution set is:{-2, i, i^2}?
    Oh no!

    Why do you reckon i^2 is a root ?
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  13. #13
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    Quote Originally Posted by Archie Meade View Post
    Oh no!

    Why do you reckon i^2 is a root ?
    Lol, stop being a bully.

    i = \sqrt{-1}
    Therefore, i^2 = -1

    What are the solutions to \sqrt{4}?

    So, what are the solutions to \sqrt{-1}? One is i. What is the other one?
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  14. #14
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    Quote Originally Posted by Quacky View Post
    Lol, stop being a bully.

    i = \sqrt{-1}
    Therefore, i^2 = -1

    What are the solutions to \sqrt{4}?

    So, what are the solutions to \sqrt{-1}? One is i. What is the other one?
    1*-1? I think?
    Am I smart now?
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  15. #15
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    the roots of the equation are x = -2 , x = i , and x = -i

    imaginary roots form conjugate pairs.
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