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Wednesday, 6 august 2008 |
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Equation of thirs grade |
Proposed by
cipriancx |
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| (25 comments) | 39.831 times displayed |
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Gasiti x numar real astfel incat :
x^3+(x+1)^3+(x+2)^3+(x+3)^3=0,
unde prin x^3 s-a notat x la puterea a 3-a |
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Ne vom folosi de dezvoltarea lui (a+b)^3 pt a scrie
a^3+b^3=(a+b)^3-3ab(a+b)=(a+b)[(a+b)^2-3ab]
Suma mai poate fi scrisa
X^3+(x+3)^3+(x+1)^3+(x+2)^3=0
(x+x+3)[(x+x+3)^2-3x(x+3)]+(x+1+x+x2)[(x+1+x+2)^2-3(x+1)(x+2)]=0
(2x+3)[ 2(2x+3)^2-3x(x+3)-3(x+1)(x+2)]=0
(2x+3)(8x^2+24x+18-3x^2-9x-3x^2-9x-2)=0
(2x+3)(2x^2+6x+16)=0
Acum simplificam prin 2
(2x+3)(x^2+3x+8)=0
Acum unul din cei doi factori este 0.Dar cum x^2+3x+8 are delta mai mic ca 0(nu avem solutie)rezulta ca 2x+3=0 => x=-3/2 |
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