The answer is x= -2/3 negative two over three.. How do we come to -2/3? Can someone please spell this out for me? Thanks. =)
2007-04-14 11:39:45 · 6 answers · asked by grem 3 in Science & Mathematics ➔ Mathematics
9x² + 12x = - 4
9x² + 12x + 4 = - 4 + 4
9x² + 12x + 4 = 0
The middle term is + 12x
Find the sum of the middle term
Multiply the first term 9 times the last term 4 equals 36 and factor
factors of 36
1 x 36
2 x 18
3 x 12
4 x 9
6 x 6. . . .<=. .use these factors
+ 6 and + 6 satisfy the sum of the middle term
Insert + 6x and + 6x into the equation
9x² + 12x - 4 = 0
9x² + 6x + 6x + 4 = 0
3x(3x + 2) + 2(3x + 2) = 0
(3x + 2)(3x + 2) = 0
- - - - - -
Roots
3x + 2 = 0
3x + 2 - 2 = 0 - 2
3x = - 2
3x/3 = - 2/3
x = - 2/3
- - - - - - - -s-
2007-04-14 13:27:26 · answer #1 · answered by SAMUEL D 7 · 2⤊ 0⤋
Move terms to left side
9x^2 + 12x + 4 = 0
Factor:
(3x + 2)(3x + 2) = 0
Set factors equal to 0 and solve
3x + 2 = 0
3x = -2
x = -2/3
2007-04-14 11:48:15 · answer #2 · answered by suesysgoddess 6 · 2⤊ 0⤋
9x² + 12x + 4 = 0
9x² + 6x + 6x + 4 = 0
3x(3x+2) + 2(3x+2) = 0
(3x+2)(3x+2) = 0
(3x+2)² = 0
3x + 2 = 0
3x = -2
x = -2/3
2007-04-14 11:50:18 · answer #3 · answered by Anonymous · 1⤊ 0⤋
9x^2+12x+4 = 0.
factor into (3x+2)(3x+2)=0
x=-2/3 is a double root
2007-04-14 11:47:34 · answer #4 · answered by David K 3 · 0⤊ 0⤋
9x^2+12x+4=0
(3x+2)(3x+2)or (3x+2)^2=0
3x+2=0
3x+2-2=0-2
3x=-2
x=-2/3
2007-04-14 12:08:31 · answer #5 · answered by Dave aka Spider Monkey 7 · 1⤊ 0⤋
use quadratic formula
x= [-b +/1 sqrt(b^2 - 4ac)] / 2a
when f(x) = ax^2 +bx+c
9x^2 + 12x + 4 = 0
x1= [- 12 + sqrt(12^2 - 4(9)4)] / 2(9) = [-12 + sqrt(0)] /18
= -12/18 = -2/3
x2 = [-12 - sqrt(0)] / 18 = -12/18 = -2/3
2007-04-14 11:51:41 · answer #6 · answered by randarinne 2 · 1⤊ 0⤋