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

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

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

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

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

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

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