solve the following equation by using the quadratic formula.
s^2 - 4s - 2 =0
2007-09-01 10:14:59 · 8 answers · asked by What's up people!! 1 in Science & Mathematics ➔ Mathematics
s=[-b±√(b²-4ac)]/2a = [4±√(16+4*1*2)]/2
=(4±√24)/2
2±√6
2007-09-01 10:22:35 · answer #1 · answered by chasrmck 6 · 0⤊ 0⤋
2
2007-09-01 17:18:26 · answer #2 · answered by Anonymous · 0⤊ 3⤋
-b±sqroot(b²-4ac)/2a
a = 1, b = -4, c = -2
-(-4)±sqroot((-4)²-4(1)(-2))/2(1)
4 ± sqroot(16+8)/2(1)
4 ± sqroot(24)/2(1)
4 ± sqroot(2*2*2*3)/2(1)
*** (2*2*2*3) = 24.. simplifying because its not a 'nice' sqroot.
4 ± 2sqroot(2*3)/2(1)
4 ± 2sqroot(2*3)/2
2 ± sqroot(6)
First answer:
2 + sqroot(6)
Second answer:
2 - sqroot(6)
2007-09-01 17:34:10 · answer #3 · answered by roOt 3 · 0⤊ 0⤋
s = (4 +/- sqrt(16 + 4*2)) / 2
= (4 +/- sqrt(24)) / 2
= (4 +/- sqrt(4)sqrt(6)) / 2
= 2 +/- sqrt(6).
2007-09-01 17:22:19 · answer #4 · answered by Anonymous · 0⤊ 0⤋
a=1
b=-4
c=-2
s=-(b +/- sqrt(b^2 - 4ac))/2a
s=4/2 +/- sqrt((16+8)/2 = 4 +/- sqrt(24)
s=2 +/- sqrt(4*6) = 2 +/-2sqrt(6)
2007-09-01 17:23:10 · answer #5 · answered by skipper 7 · 0⤊ 1⤋
I think it's 2 (plus or minus) 2i (squareroot) of 2
Looking at the other answers...I messed something up...
2007-09-01 17:26:24 · answer #6 · answered by ? 2 · 0⤊ 0⤋
s = [4+/- sqrt(4^2-4(1)(-2)]/2
s = [4 +/- sqrt(24)]/2
s = [4 +/- 2sqrt(6)]/2
s = 2 +/- sqrt(6)
2007-09-01 17:24:16 · answer #7 · answered by ironduke8159 7 · 0⤊ 0⤋
Rackbrane is correct!
2007-09-01 17:26:08 · answer #8 · answered by Ken H 1 · 0⤊ 0⤋