2006-06-28 17:14:44 · 6 answers · asked by Jonathan T 1 in Science & Mathematics ➔ Mathematics
3x^2 - 4x = 2x^2 + 2
x^2 - 4x - 2 = 0
x = (-b ± sqrt(b^2 - 4ac))/2a
x = (-(-4) ± sqrt((-4)^2 - 4(1)(-2)))/(2(1))
x = (4 ± sqrt(16 + 8))/2
x = (4 ± sqrt(24))/2
x = (4 ± sqrt(4 * 6))/2
x = (4 ± 2sqrt(6))/2
x = 2 ± sqrt(6)
ANS : 2 ± sqrt(6)
2006-06-28 17:19:54 · answer #1 · answered by Sherman81 6 · 0⤊ 0⤋
3x^2-4x=2x^2+2
Ans:
=>3x^2-4x-2x^2-2=0
=>x^2-4x-2=0
x=(4-(16+8)^(1/2))/2
x=(4-4.899)/2
x1=.4495
x2=4.4495
2006-06-28 23:10:11 · answer #2 · answered by chella 1 · 0⤊ 0⤋
The answers are:
X1= 4.45
X2 = -0.45
You can solve using the quadratic equation after transposing every term to one side and equating to zero.
2006-06-28 19:04:42 · answer #3 · answered by boo0726 3 · 0⤊ 0⤋
2 + 2^(.5)
or
2 - 2^(.5)
2006-06-28 17:22:08 · answer #4 · answered by dlfield 3 · 0⤊ 0⤋
2+sqrt(6)=4.449489743
or
2-sqrt(6)=-0.449489743(minus)
2006-06-28 23:17:48 · answer #5 · answered by Piyush 2 · 0⤊ 0⤋
use ur calculator.....
the answer will be
1.1483314
2006-06-28 17:47:40 · answer #6 · answered by Black Eye Peace 2 · 0⤊ 0⤋