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Thread: Seeking Math help +rep pre cal questions

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    Seeking Math help +rep pre cal questions

    1. Translate the statement below to an equation using absolute value and solve:
    The distance between 50 and me is at least 10.


    2. Using the given zero, find all of the zeros and write a linear factorization of f(x).

    1+3i is a zero of f(x)=x^4-2x^3+5x^2+10x-50

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    1. Translate the statement below to an equation using absolute value and solve:
    The distance between 50 and me is at least 10.

    |x-50| ≥ 10

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    thanks homie

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    No problem. I don't remember how to do #2 though D:

    ---------- Post added at 03:32 PM ---------- Previous post was at 03:23 PM ----------

    http://www.wolframalpha.com/input/?i...x%5E2%2B10x-50

  5. The Following User Says Thank You to Carrot For This Useful Post:

    John (01-31-2012)

  6. #5
    bamag's Avatar
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    1+3i is a zero of f(x)=x^4-2x^3+5x^2+10x-50
    Since 1+3i is a root, 1-3i is also a root. ((x-1)-3i)((x-1)+3i)=(x-1)^2-(3i)^2=x^2-2x+1-9i^2=x^2-2x+10
    Use long division to get x^2-5.
    Find the two roots of that are +- radical 5.
    Therefore, all the zeroes are 1+3i,1-3i, radical 5, negative radical 5.
    Linear factorization is f(x)=(x-1-3i)(x-1+3i)(x+5^.5)(x-5^.5)
    Last edited by bamag; 01-31-2012 at 03:47 PM.

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    Quote Originally Posted by bamag View Post
    1+3i is a zero of f(x)=x^4-2x^3+5x^2+10x-50
    Since 1+3i is a root, 1-3i is also a root. ((x-1)-3i)((x-1)+3i)=(x-1)^2-(3i)^2=x^2-2x+1-9i^2=x^2-2x+10
    Use long division to get x^2-5.
    Find the two roots of that are +- radical 5.
    Therefore, all the zeroes are 1+3i,1-3i, radical 5, negative radical 5.
    Linear factorization is f(x)=(x-1-3i)(x-1+3i)(x+5^.5)(x-5^.5)


    it is cool, but you may give the explanation that why 1-3i is also a root.


    you may add [since all coefficient are real number, therefore (a+bi)(1+3i)=constant +0i,hence a=1,b=3 ]

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    Quote Originally Posted by dodowong303 View Post
    it is cool, but you may give the explanation that why 1-3i is also a root.


    you may add [since all coefficient are real number, therefore (a+bi)(1+3i)=constant +0i,hence a=1,b=3 ]
    It is simply the conjugate roots theorem. If a+bi is a root, then a-bi is also a root, and vise versa.

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    Quote Originally Posted by bamag View Post
    It is simply the conjugate roots theorem. If a+bi is a root, then a-bi is also a root, and vise versa.
    i know but my teacher will deduct marks form missing some stupid simple explanation such as a+bi&a-bi must be pair,etc.......==

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