2x^2-4x-3=0
please help
2007-06-30 02:54:02 · 8 answers · asked by Bianca B 1 in Science & Mathematics ➔ Mathematics
2x^2-4x-3=0
2(x^2-2x)-3=0
2(x-1)^2-5=0
(x-1)^2=5/2
x-1= √(5/2)
x-1=[(√10)/2]
x=±[(√10)/2]+1
2007-06-30 03:06:05 · answer #1 · answered by Popo B 3 · 1⤊ 0⤋
x² - 2x = 3/2
x² - 2x + 1 = 3/2 + 1
(x - 1)² = 5 / 2
(x - 1) = 屉(5/2)
x = 1 ± â(5/2)
2007-06-30 04:22:28 · answer #2 · answered by Como 7 · 0⤊ 0⤋
2 x^2 - 4 x - 3 = 0
x^2 - 2 x - 3/2 = 0 . . . . . . divided by 2
x^2 - 2 x + 1 - 1 - 3/2 = 0
( x - 1 )^2 - 5/2 = 0
2007-06-30 03:03:46 · answer #3 · answered by CPUcate 6 · 0⤊ 0⤋
hey dude here's d solution:
2x^2-4x-3=0 4/2=(2)^2=4
(2x^2-4x)=3
(2x^2-4x+4)=3+4
(2x^2-4x+4)=7
2007-06-30 02:56:45 · answer #4 · answered by Anonymous · 0⤊ 1⤋
2x^2-4x-3=0
divide both sides by 2;
x^2-2x=3/2
add1 to both sids;
x^2-2x+1=3/2+1;
(x-1)^2=5/2=2.5
x-1=+/-1.58=+2.58,-0.58answer
2007-06-30 03:30:49 · answer #5 · answered by Anonymous · 0⤊ 0⤋
X^2 -2X = 3/2
X^2 - 2X + 1 = 3/2 + 1
(X -1)^2 = 5/2
X -1 = +/- (SQ RT (5/2)) = +/- (SQ RT10) / 2
X = 1 +/- (SQ RT 10) / 2
2007-06-30 03:17:26 · answer #6 · answered by spirit dummy 5 · 0⤊ 0⤋
x1 = 1 + 2 * sqare root of 10
x2 = 1 - 2 * sqare root of 10
2007-06-30 03:07:47 · answer #7 · answered by Anonymous · 0⤊ 0⤋
2x^2-4x-3
2(x^2-2x+1)-2-3=0
Or (x-1)^2=-5/2
or x-1=(+-)(5/2)^1/2
Or x=(+-)(5/2)^1/2+1
2007-06-30 10:15:58 · answer #8 · answered by MAHAANIM07 4 · 0⤊ 0⤋