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Math Help - Need help with a few equations

  1. #1
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    Need help with a few equations

    Hey,
    surely those exercises are easy for you guys, but I need them to get done by thursday. I missed a few lessons, so I really don't know how to do these equations.


    x^3-3x^2+x+5=0

    2x^4-9x^3+4x^2+21x-18=0 - Using the Horner scheme

    Help would be appreciated.
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  2. #2
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by Rix12 View Post
    Hey,
    surely those exercises are easy for you guys, but I need them to get done by thursday. I missed a few lessons, so I really don't know how to do these equations.


    x^3-3x^2+x+5=0

    2x^4-9x^3+4x^2+21x-18=0 - Using the Horner scheme

    Help would be appreciated.
    I don't know the Horner scheme so I shall do the first question


    From inspection I know that f(-1) = 0

    Thus using the factor theorem (see spoiler below for what it is) I know that (x+1) is a factor

    Spoiler:
    The factor theorem states that q(x) is a factor of f(x) if, and only if, f(x-q) =0


    Which gives us x^3-3x^2+x+5 = (x+1)(Ax^2+Bx+C) where A, B and C are constants to be found.

    Comparing the coefficients:

    • x^3 \: \rightarrow \: 1 = A
    • x^0 \: \rightarrow \: 5 = C
    • x^1 \: \rightarrow \: 1 = B+C


    Thus A=1 , B = -4 and C=5

    Put this back into our equation: (x+1)(x^2-4x+5)

    Now we must check if the quadratic term will factor (ie if the discriminant is a square number or not): \Delta = b^2-4ac = (-4)^2-4 \cdot 1 \cdot 5 = -4.

    This will not give a real solution so the final answer is (x+1)(x^2-4x+5)=0 thus x=-1 in the real numbers


    Edit: Oops, forgot to see it was equal to 0 - complex solutions upcoming

    x^2-4x+5 = 0

    From the quadratic formula: x = \dfrac{4 \pm \sqrt{(-4)^2-4 \cdot 1 \cdot 5}}{2} = \dfrac{4 \pm \sqrt{-4}}{2} \: = \dfrac{4\pm 2i}{2} = 2 \pm i

    where i^2 = -1


    Thus all the solutions are: x = -1, \: 2+i, \: 2-i
    Last edited by e^(i*pi); September 18th 2010 at 10:09 AM. Reason: Added complex numbers
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  3. #3
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    Thank's a lot mate. Got the idea now.

    Actually there's one more I forgot to ask:

    Convert this one to algebraic form:
    e^[(-1)-2*i]
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  4. #4
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    Today I learned that Horner method not only computes the value of a polynomial f(x) at a given point x=a but also gives the ratio and the remainder of f(x) and x-a.

    Let f_1(x)=2x^4-9x^3+4x^2+21x-18. We can guess that x_1=1 is a root. (See below about how to make an educated guess.) Dividing f_1(x) by x-1 gives f_2(x)=2x^3-7x^2-3x+18. (Instead of describing this division, I'll describe the next one.) Again by guessing, x_2=2 is a root of f_2(x).

    To divide f_2(x) by x-2, make a table with three rows. The first row consists of the coefficients of f_2. The first element of the second row is 0. The third row is the sum of the first two. Therefore, the first element of the third row is 2.

    Code:
     2 -7 -3  18
     0  4 -6 -18
     2 -3 -9   0
    The second and each subsequent element of the second row is x_2=2 times the element of the third row immediately to the left. So, 4 = 2 * 2, -6 = 2 * (-3), and -18 = 2 * (-9). One computes the second and third row simultaneously left to right. The result indicates that the ratio is the polynomial f_3(x)=2x^2-3x-9 and the remainder is 0. Now, f_3(x)=0 is a quadratic equation.

    To make an educated guess about the roots, one can use the rational root theorem. It says that if a rational number p/q (in the lowest terms) is a root of f(x), then p divides the constant term of f(x) and q divides the leading coefficient. So, for f_2(x), all rational roots are among \pm1, \pm2, \pm 3, \pm 6, \pm 9, \pm 18, \pm 1/2, \pm 3/2, \pm 9/2. Note that the theorem says nothing about irrational roots.
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  5. #5
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    Quote Originally Posted by Rix12 View Post
    Thank's a lot mate. Got the idea now.

    Actually there's one more I forgot to ask:

    Convert this one to algebraic form:
    e^[(-1)-2*i]
    You are clearly expected to know two things in order to be able to answer this question:

    1) e^{a+ b}= e^{a}e^{b} and

    2) e^{ix}= cos(x)+ i sin(x).
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  6. #6
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    Thanks guys. That helped a lot.
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