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nickersia
 one year ago
Medal to first who explains me this question :)
Find the roots of the equation (comment).
nickersia
 one year ago
Medal to first who explains me this question :) Find the roots of the equation (comment).

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nickersia
 one year ago
Best ResponseYou've already chosen the best response.0\[x ^{3}x ^{2}7x+3\]

shelbygt520
 one year ago
Best ResponseYou've already chosen the best response.1k whats the question

shelbygt520
 one year ago
Best ResponseYou've already chosen the best response.1ok do you want it factorized or what?

nickersia
 one year ago
Best ResponseYou've already chosen the best response.0That's problem, I'm not quite sure what should I do :s

shelbygt520
 one year ago
Best ResponseYou've already chosen the best response.1x^3+x^27x3 One of the roots is x(1) = 3 After applying Polynomial remainder theorem you will have a second order equation x^2  2x  1 = 0 a = 1 ; b = 2 ; c = 1 x(2,3) = (b)(+/)SqRoot{b^2  4 . a . c}/(2 a)= x(2,3) = (2)(+/)SqRoot{4  4 . 1 . (1)}/(2 . 1)= x(2,3) = (2)(+/)SqRoot{4 + 4}/(2)= x(2,3) = 2 (+/) SqRoot { 8 } / 2 = x(2,3) = 1 (+/) SqRoot { 2 } > x(2) = 1 + SqRoot { 2 } > x(3) = 1  SqRoot { 2 }

shelbygt520
 one year ago
Best ResponseYou've already chosen the best response.1u should do that but instead of doing minus you do the addition this was a mere example but close enough so that you would know what to do

nickersia
 one year ago
Best ResponseYou've already chosen the best response.0I understand second part perfectly that's not problem :) can you just explain how did you get 3 as one of the roots? I'm bit confused

shelbygt520
 one year ago
Best ResponseYou've already chosen the best response.1I just replaced +3 with 3 in this example

shelbygt520
 one year ago
Best ResponseYou've already chosen the best response.1so in your case the root would be +3
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