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Thread: A small doubt

  1. #1
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    A small doubt

    Is there any way to find sq.root of a fourth degree polynomial?
    In this question, x^4 -2x^3 +3x^2 -2x +1 ,how can we find. Sq.root?



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  2. #2
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    Re: A small doubt

    I would write:

    $\displaystyle x^4-2x^3+3x^2-2x+1=(x^2+ax+b)^2=x^4+2ax^3+(a^2+2b)x^2+2abx+b^2$

    Equating coefficients, we obtain the system:

    $\displaystyle 2a=-2\implies a=-1$

    $\displaystyle a^2+2b=3\implies b=1$

    And so we have:

    $\displaystyle x^4-2x^3+3x^2-2x+1=(x^2-x+1)^2$

    What do you then conclude?
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    Re: A small doubt

    Quote Originally Posted by MarkFL View Post
    I would write:

    $\displaystyle x^4-2x^3+3x^2-2x+1=(x^2+ax+b)^2=x^4+2ax^3+(a^2+2b)x^2+2abx+b^2$

    Equating coefficients, we obtain the system:

    $\displaystyle 2a=-2\implies a=-1$

    $\displaystyle a^2+2b=3\implies b=1$

    And so we have:

    $\displaystyle x^4-2x^3+3x^2-2x+1=(x^2-x+1)^2$

    What do you then conclude?
    Thats great. So we should use the principle that square of n termed polynomial have 2n terms.And then equate..
    Thanks again bro..


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    Re: A small doubt

    Quote Originally Posted by MarkFL View Post
    I would write:

    $\displaystyle x^4-2x^3+3x^2-2x+1=(x^2+ax+b)^2=x^4+2ax^3+(a^2+2b)x^2+2abx+b^2$

    Although it's natural to do the above, and I would expect someone to set it up that way, it could also be done this way:


    You could write:

    $\displaystyle x^4 - 2x^3 + 3x^2 - 2x + 1 \ = \ (-x^2 + ax + b)^2 \ = \ x^4 - 2ax^3 + (a^2 - 2b)x^2 + 2abx + b^2$


    Equating coefficients, we obtain the system:

    $\displaystyle -2a \ = \ -2 \ \implies \ a \ = \ 1$

    $\displaystyle a^2 - 2b \ = \ 3 \ \implies \ b \ = \ -1$

    And so we have:

    $\displaystyle x^4 - 2x^3 + 3x^2 - 2x + 1 \ = \ (-x^2 + x - 1)^2$


    __________________________________________________ _______________________


    sbjsavio, it is not true in general that the square of an n-termed polynomial (restricting ourselves to one variable for this discussion)
    has 2n terms (when the like terms have been combined).

    Look at these counterexamples:


    $\displaystyle (x)^2 \ = \ x^2$

    $\displaystyle (x + 1)^2 \ = \ x^2 + 2x + 1$

    $\displaystyle (x^2 + 4x - 8)^2 \ = \ x^4 + 8x^3 - 64x + 64$


    __________________________________________________ ________


    Quote Originally Posted by sbjsavio View Post
    Is there any way to find sq.root of a fourth degree polynomial?
    In this question, x^4 -2x^3 +3x^2 -2x +1 ,how can we find. Sq.root?
    Also, there is a square root algorithm for polynomials that someone gives as an example here:

    https://math.stackexchange.com/quest...f-a-polynomial
    Last edited by greg1313; Jul 11th 2018 at 07:23 AM.
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  5. #5
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    Re: A small doubt

    Quote Originally Posted by greg1313 View Post
    Although it's natural to do the above, and I would expect someone to set it up that way, it could also be done this way:


    You could write:

    $\displaystyle x^4 - 2x^3 + 3x^2 - 2x + 1 \ = \ (-x^2 + ax + b)^2 \ = \ x^4 - 2ax^3 + (a^2 - 2b)x^2 + 2abx + b^2$


    Equating coefficients, we obtain the system:

    $\displaystyle -2a \ = \ -2 \ \implies \ a \ = \ 1$

    $\displaystyle a^2 - 2b \ = \ 3 \ \implies \ b \ = \ -1$

    And so we have:

    $\displaystyle x^4 - 2x^3 + 3x^2 - 2x + 1 \ = \ (-x^2 + x - 1)^2$


    __________________________________________________ _______________________


    sbjsavio, it is not true in general that the square of an n-termed polynomial (restricting ourselves to one variable for this discussion)
    has 2n terms (when the like terms have been combined).

    Look at these counterexamples:


    $\displaystyle (x)^2 \ = \ x^2$

    $\displaystyle (x + 1)^2 \ = \ x^2 + 2x + 1$

    $\displaystyle (x^2 + 4x - 8)^2 \ = \ x^4 + 8x^3 - 64x + 64$


    __________________________________________________ ________




    Also, there is a square root algorithm for polynomials that someone gives as an example here:

    https://math.stackexchange.com/quest...f-a-polynomial
    Oh,thanks greg ,this is easier


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    Re: A small doubt

    Quote Originally Posted by sbjsavio View Post
    x^4 -2x^3 +3x^2 -2x +1 , how can we find. Sq.root?
    On a timed test, with about a minute left, you'd go this way:
    a^2 = x^4 -2x^3 +3x^2 -2x +1
    a = sqrt(x^4 -2x^3 +3x^2 -2x +1)

    Well, perhaps a point or two
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    Re: A small doubt

    Quote Originally Posted by DenisB View Post
    On a timed test, with about a minute left, you'd go this way:
    a^2 = x^4 -2x^3 +3x^2 -2x +1
    a = sqrt(x^4 -2x^3 +3x^2 -2x +1)

    Well, perhaps a point or two
    Yup ..


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