2007-06-30 04:56:58 · 4 answers · asked by CPUcate 6 in Science & Mathematics ➔ Mathematics
x = [ -2 +- sqr(3) ] /2
2x = [ -2 +- sqr(3) ]
2x+2 = +/- sqrt(3)
(2x+2)^2 = 3
4x^2 +8x +4 = 3
4x^2 +8x+1 = 0
2007-06-30 05:20:21 · answer #1 · answered by ironduke8159 7 · 0⤊ 0⤋
x^2 - ( sum of the two roots ) x + product of the roots = 0
x^2 - ([ -2 + sqr(3) ] /2 + [ -2 - sqr(3) ] /2) x+ ([ -2 + sqr(3) ] /2 ) ( -2 - sqr(3) ] /2)=0
x^2 + 2 x+ 1/2 =0
2x^2 + 4x+ 1 =0
2007-06-30 12:16:57 · answer #2 · answered by pioneers 5 · 0⤊ 1⤋
x= -2 +/- sqr(3)
x+2 = +/- sqr (3)
(x+2) ^ 2 = 3
X^2 +4X + 4 = 3
X^2 +4X + 1 = 0
2007-06-30 12:03:24 · answer #3 · answered by telsaar 4 · 1⤊ 0⤋
X= ( -2 +SQRT(3))/2
X - ( -2 +SQRT(3))/2 =0
X - ( -2 -SQRT(3))/2 =0
(X - ( -2 +SQRT(3))/2 ) * (X - ( -2 -SQRT(3))/2 )= X^2+2*X+1/4
2007-06-30 13:34:47 · answer #4 · answered by ? 5 · 0⤊ 0⤋