2007-04-21 12:20:41 · 7 answers · asked by robert v 1 in Science & Mathematics ➔ Mathematics
6(x - 3)² - 12 = 0
Add 12 to both sides:
6(x - 3)² = 12
Divide both sides by 6:
(x - 3)² = 2
Take the square root of both sides:
(x - 3) = ±√2
Add 3 to both sides:
x = 3 ±√2
2007-04-21 12:24:48 · answer #1 · answered by عبد الله (ドラゴン) 5 · 0⤊ 0⤋
6(x-3)^2 - 12 = 0 Simplify by dividing by 6
(x-3)^ - 2 = 0 add 2 to each side
(x - 3)^2 = 2 take the square root of each side
x - 3 = +/- (2)^1/2
x = 3 +/- Square root of 2
2007-04-21 12:41:03 · answer #2 · answered by kale_ewart 5 · 0⤊ 0⤋
6(x^2-6x+9)-12=0
6x^2-36x+54-12=0
6x^2-36x+42=0
use the quadratic formula: [-b+sq. root of(+/-b^2-4ac)] / 2a
6x^2-36x+42=0 which is ax^2+bx-c
where a=6, b=-36 and c= 42
u will get 2 answers for x, 1 positive and 1 negative.
2007-04-21 13:15:01 · answer #3 · answered by misa 2 · 0⤊ 0⤋
6x^2-18^2-12=0
6x^2-18^2=12
6x^2-324+324=12+324
6x^2/6=336/6
x^2=56
x=sq root 56
x=7.48
I Think.......
2007-04-21 12:42:27 · answer #4 · answered by Jordan W 1 · 0⤊ 0⤋
6(x^2-6x+9)-12=0
6x^2-36x+54-12=0
6x^2-36x+42=0
6(x^2-6x+7)=0
6(x-1)(x-6)=0
x=1or 6
2007-04-21 12:50:08 · answer #5 · answered by Dave aka Spider Monkey 7 · 0⤊ 0⤋
(6x -18)(-10)= 0
-60x+180=0
-60x=-180
x=-180/-60
x=3
2007-04-21 12:29:51 · answer #6 · answered by Rob 2 · 0⤊ 1⤋
6(x-3)^2-12=0
6(x-3)^2=12
6(x-3)(x-3)=12
6(x^2-6x+9)=12
6x^2-36x+54=12
6x^2-36x+54-12=0
6x^2-36x+42=0
6(x^2-6+7)=0
2007-04-21 12:44:56 · answer #7 · answered by bootis32 6 · 0⤊ 0⤋