x^2-3x+7
2007-11-21 03:29:46 · 7 answers · asked by lele 2 in Science & Mathematics ➔ Mathematics
because b^2-(4*a*c)<0 you have 2 solutions which are complex. (a=2, b=3, c=7).
so.. the solutions are:
x1=(3+19i)/2
x2=(3-19i)/2. where i=sqrt(-1)
2007-11-21 03:37:03 · answer #1 · answered by nobody100 4 · 0⤊ 1⤋
split the middle term in such a way that the sum of the coefficients is 3 and the product is 7.
PS:IT MIGHT NOT WORK.
2007-11-21 03:47:47 · answer #2 · answered by Anonymous · 0⤊ 0⤋
x = -b + or - sqrt(b^2 -4ac)/2
a = 1 b = -3 and c = 7
2007-11-21 03:44:00 · answer #3 · answered by ikeman32 6 · 0⤊ 0⤋
x^2 - 3x + 7 = 0
complete the square
x^2 - 3x + 2.25 = -7 + 2.25
(x-1.5)^2 = -4.75
x - 1.5 = sqrt(-4.75) = sqrt (-19/4) = sqrt(-19)/2
x = 1.5 +/- .5i*sqrt(19)
2007-11-21 03:44:03 · answer #4 · answered by Steve A 7 · 0⤊ 0⤋
You have to use the "completing the square method"
Hope that helps.
2007-11-21 03:41:15 · answer #5 · answered by why do you care? 3 · 0⤊ 0⤋
hate to be a pest, but is it equal to anything?
2007-11-21 03:44:00 · answer #6 · answered by Anonymous · 0⤊ 1⤋
4x-5
2007-11-21 03:37:23 · answer #7 · answered by Anonymous · 0⤊ 2⤋