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Zeros of a polynomial

  • Thread starter ehrenfest
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  • #1
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[SOLVED] zeros of a polynomial

Homework Statement


Find the zeros of the polynomial

[tex]P(x) = x^4-6x^3+18x^2-30x+25[/tex]

knowing that the sum of two of them is 4.


Homework Equations


http://en.wikipedia.org/wiki/Viète's_formulas


The Attempt at a Solution


Let x_1,x_2,x_3,x_4 be the complex roots and let x_1 +x_2 = 4. Here are the Viete relations in this case:

[tex]x_1+x_2+x_3+x_4 = 6 [/tex]

[tex] x_1 x_2 +x_1 x_3 +x_1 x_4 + x_2 x_3 + x_2 x_4 +x_3 x_4 = 18 [/tex]

[tex] x_1 x_2 x_3 + x_1 x_3 x_4 +x_2 x_3 x_4 +x_1 x_2 x_4= 30 [/tex]

[tex] x_1 x_2 x_3 x_4 = 25[/tex]

The first one implies that x_3 +x_4 =2. And then the second one implies that x_1 x_2 + x_3 x_4 = 10 but that is as far as I can get.

Please just give a hint.
 

Answers and Replies

  • #2
tiny-tim
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x_1 +x_2 = 4 … x_3 +x_4 =2.
Hint: so x_2 = 4 - x_1 and x_4 = 2 - x_3.

Go substitute! :smile:
 
  • #3
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Do you mean into

[tex]
P(x) = x^4-6x^3+18x^2-30x+25
[/tex]

?
 
  • #4
tiny-tim
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No … I mean substitute into your Viete relations. :smile:
 
  • #5
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Sorry tiny-tim, I don't know see why that helps:

For the second Viete relation I get:

[tex]4x_1 - 2 x_3 - x_1^2-x_3 ^2+ x_1 x_3 = 10[/tex]

For the third one I get

[tex]8x_1 + 8x_3 - 2 x_1 ^2 - 4 x_3^2 = 30[/tex]

For the fourth one I get

[tex]8 x_1 x_3 + x_1 ^2 x_ 3^2 - 2x_1 ^2 x_3 - 4 x_1 x_3 ^2 = 25[/tex]

Is there a simple way to solve these equations?
 
  • #6
tiny-tim
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hmm … turned out more complicated than I thought. :frown:

Well … that's what happens in exams sometimes … you try something, and it doesn't work, so you try the next most obvious thing … :smile:

Now this will work:

in your
[tex] x_1 x_2 +x_1 x_3 +x_1 x_4 + x_2 x_3 + x_2 x_4 +x_3 x_4 = 18 [/tex]

[tex] x_1 x_2 x_3 + x_1 x_3 x_4 +x_2 x_3 x_4 +x_1 x_2 x_4= 30 [/tex]
put (x_1 + x_2)s together, and (x_3 + x_4)s, and you should get two equations in x_1x_2 and x_3x_4, from which you get x_1x_2 = … and x_3x_4 = … ? :smile:
 

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