2006-08-13 18:48:17 · 5 answers · asked by t13fool 1 in Science & Mathematics ➔ Mathematics
multiply the two side by x², because 0 is not a solution :
3x²-2x+1 =0
So
D= 4-12 = -8 = -2* 4
No solutions in real only in imaginary
Solutions
x1= (1 + sqrt(2) i)/3
x2 = (1- sqrt(2) i)/3
2006-08-13 18:59:30 · answer #1 · answered by fred 055 4 · 2⤊ 0⤋
No. As the other guy says there is no real solution. But for sure it will not be 0
2006-08-13 18:59:18 · answer #2 · answered by canguy71 2 · 0⤊ 0⤋
3-(2/x)+(1/x^2)=0
Multiply both sides by by x^2
3x^2 - 2x + 1 = 0
Quadratic equation
(2 +/- sqrt(4 - (4)(3)(1))) / 2(3)
(2 +/- sqrt(4 - 12)) / 6
There is no real solution because 4 - 12 < 0
2006-08-13 18:52:53 · answer #3 · answered by Michael M 6 · 0⤊ 0⤋
no real solution
2006-08-13 19:18:07 · answer #4 · answered by Xavi DK 3 · 0⤊ 0⤋
Sure, why not?
2006-08-13 18:51:14 · answer #5 · answered by Anonymous · 0⤊ 0⤋