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Math Help - working backwards - cubics

  1. #1
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    working backwards - cubics

    Write an equation that has the following roots: 2, -1, 5

    Answer key: x^3 - 6x^2 + 3x + 10 = 0

    For quadratic equations, I use the sum and product of roots, this is a cubic equation, how do I solve this?

    Thanks.
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  2. #2
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    Quote Originally Posted by shenton View Post
    Write an equation that has the following roots: 2, -1, 5

    Answer key: x^3 - 6x^2 + 3x + 10 = 0

    For quadratic equations, I use the sum and product of roots, this is a cubic equation, how do I solve this?

    Thanks.
    (x - 2)(x + 1)(x - 5)
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  3. #3
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    Thanks! That turns out to be not as difficult as imagined. I thought I needed to use sum and products of roots to write the equation, it does makes me wonder a bit why or when I need to use sum and products of roots.
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  4. #4
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    Question

    Write an equation that has the following roots: 2, -1, 5

    Is there any other way to solve this other than the (x-2)(x+1)(x-5) method?

    If we have these roots: 1, 1 + √2, 1 - √2

    the (x - 1) (x -1 -√2) (x -1 +√2) method seems a bit lenghty.

    When we expand (x - 1) (x -1 -√2) (x -1 +√2) the first 2 factors,

    it becomes:

    (x^2 -x -x√2 -x +1 +√2) (x -1 +√2)

    collect like terms:

    (x^2 -2x -x√2 +1 +√2) (x -1 +√2)

    To further expand this will be lenghty, my gut feel is that mathematicians do not want to do this - it is time consuming and prone to error. There must be a way to write an equation other than the above method.

    Is there a method to write an equation with 3 given roots (other than the above method)?

    Thanks.
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  5. #5
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    Quote Originally Posted by shenton View Post
    Write an equation that has the following roots: 2, -1, 5

    Is there any other way to solve this other than the (x-2)(x+1)(x-5) method?

    If we have these roots: 1, 1 + √2, 1 - √2

    the (x - 1) (x -1 -√2) (x -1 +√2) method seems a bit lenghty.

    When we expand (x - 1) (x -1 -√2) (x -1 +√2) the first 2 factors,

    it becomes:

    (x^2 -x -x√2 -x +1 +√2) (x -1 +√2)

    collect like terms:

    (x^2 -2x -x√2 +1 +√2) (x -1 +√2)

    To further expand this will be lenghty, my gut feel is that mathematicians do not want to do this - it is time consuming and prone to error. There must be a way to write an equation other than the above method.

    Is there a method to write an equation with 3 given roots (other than the above method)?

    Thanks.
    You have a pair of roots of the form a+sqrt(b) and a-sqrt(b) if you multiply
    the factors corresponding to these first you get:

    (x-a-sqrt(b))(x-a+sqrt(b))=x^2+(-a-sqrt(b))x+(-a+sqrt(b))x +(-a-sqrt(b))(-a+sqrt(b))

    ................=x^2 - 2a x + (a^2-b)

    Which leaves you with the easier final step of computing:

    (x-1)(x^2 - 2a x + (a^2-b))

    RonL
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  6. #6
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    Hello, shenton!

    The sum and product of roots works well for quadratic equations.

    For higher-degree equations, there is a generalization we can use.

    To make it simple (for me), I'll explain a fourth-degree equation.


    Divide through by the leading coefficient: . x^4 + Px^3 + Qx^2 + Rx + S \:=\:0

    Insert alternating signs: . +\:x^4 - Px^3 + Qx^2 - Rx + S \:=\:0
    . . . . . . . . . . . . . . . . . . \uparrow\quad\;\; \uparrow\qquad\;\;\uparrow\qquad\;\,\uparrow\qquad  \,\uparrow


    Suppose the four roots are: a,\,b,\,c,\,d.

    The sum of the roots (taken one at a time) is: -P.
    . . a + b + c + d \:=\:-P

    The sum of the roots (taken two at a time) is: Q.
    . . ab + ac + ad + bc + bd + cd \:=\:Q

    The sum of the roots (taken three at a time) is: -R.
    . . abc + abd + acd + bcd \:=\:-R

    The sum of the roots ("taken four at a time") is: S.
    . . abcd \:=\:S

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    For your problem with roots: (a,b,c) \:=\:(2,-1,5)
    . . we have: . x^3 + Px^2 + Qx + R \:=\:0

    Then: . a + b + c \:=\:-P\quad\Rightarrow\quad2 + (-1) + 5\:=-P
    . . Hence: P = -6

    And: . ab + bc + ac \:=\:Q\quad\Rightarrow\quad(2)(-1) + (-1)(5) + (2)(5) \:=\:Q
    . . Hence:  Q = 3

    And: . abc \:=\:-R\quad\Rightarrow\quad(2)(-1)(5)\:=\:-R
    . . Hence: R = 10


    Therefore, the cubic is: . x^3 - 6x^2 + 3x + 10 \:=\:0

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  7. #7
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    This is awesome, Soroban. Using the method you shown, I was able to solve this problem:

    1, 1+√2, 1-√2

    a=1, b=1+√2, c=1-√2

    Let x^3 - px^2 + qx - r = 0 be the cubic equation

    p = a + b + c
    = (1) + (1 + √2) + (1 - √2)
    = 3

    q = ab + bc + ac
    = (1)(1 + √2) + (1 + √2)(1 - √2) + (1)(1 - √2)
    = 1 + √2 + 1 - 2 + 1 - √2
    = 1

    r = abc
    = (1)(1 + √2)(1 - √2)
    = 1-2
    = -1

    Therefore x^3 - px^2 + qx - r = 0 becomes
    x^3 - 3x^2 + x - (-1) = 0
    x^3 - 3x^2 + x + 1 = 0

    Thanks for the help and detailed workings.
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