(x+5)(x+1)=12
2006-11-17 08:52:06 · 10 answers · asked by brunette 1 in Science & Mathematics ➔ Mathematics
x= -7, x= 1
x^2+6x+5=12
x^2+6x-7=0
(x+7)(x-1)=0
x+7=0 x= -7
x-1=0 x=1
2006-11-17 08:54:54 · answer #1 · answered by ? 3 · 0⤊ 0⤋
Ok First multiply out the brackets
x^2 +5x + x + 5 = 12
Subtract 12 from both sides
x^2 +5x +x +5 -12 = 0
simplify
x^2 +6x - 7 - 0
Re-factorise
(x + 7)(x - `1) = 0
Now, one of the above brackets must = 0, because anythjing multiplied by nothing = 0
so either x+7 = 0 or x-1=0
So if X+7 = 0 Then x = -7
Or if X-1 = 0 Then x = 1
so x= -7 or 1
hope that helps.
2006-11-17 16:57:11 · answer #2 · answered by John W 2 · 1⤊ 0⤋
well (x+5)(x+1) = 12
x^2 +6x +5 = 12
x^2 plus 6x = 7
and x = 1 because 1^2 is 1 plus 6(1) = 6 and 1 plus 6 = 7
2006-11-17 16:55:30 · answer #3 · answered by Redskins5435 1 · 0⤊ 0⤋
(x+5)(x+1) = x^2 + 6x + 5 = 12
x^2 + 6x - 7 = 0
(x + 7)(x - 1) = 0
x = -7, 1
2006-11-17 16:56:12 · answer #4 · answered by banjuja58 4 · 0⤊ 0⤋
x = 1 or -7
2006-11-17 16:54:09 · answer #5 · answered by drizzttownz 2 · 0⤊ 0⤋
x=3
2006-11-17 16:53:59 · answer #6 · answered by Anonymous · 0⤊ 0⤋
(x+5)(x+1) = 12
x(x+1) + 5(x+1) = 12
x^2 + x + 5x + 5 = 12
x^2 + 6x + 5 = 12
x^2 + 6x + 5 - 12 = 0
x^2 + 6x - 7 = 0
x^2 -x + 7x - 7 = 0
x(x - 1) + 7(x - 1) = 0
(x + 7)(x - 1) = 0
Either x + 7 = 0
or x - 1 = 0
Put x + 7 = 0
x = -7
Put x - 1 = 0
x = 1
The two possible values of 'x' are: 1 and -7
2006-11-19 09:10:33 · answer #7 · answered by Akilesh - Internet Undertaker 7 · 0⤊ 1⤋
(x+5)(x+1)=12
5x+1x+5+x(squared)=12
11x + x(squared) = 12
x = 1
2006-11-17 16:57:51 · answer #8 · answered by Wondergirl 2 · 0⤊ 0⤋
(x + 5)(x + 1) = 12
x2 + 5x + 1x + 5 = 12
x2 + 6x +5 = 12
x2 + 6x = 7 or
x2 + 6x -7
2006-11-17 16:54:38 · answer #9 · answered by chumbawumba91501 3 · 0⤊ 0⤋
(x+5)(x+1)=12
x*2 +6x +5=12
x*2 + 6x -7=0
x= {-b +/-Sq root( b*2-4ac)} /2a
x={ -6 +/- sq root (36 + 28) } /2a
Two roots
x= (-6 +8)/2 = +1
x= (-6 -8) /2 = -7
2006-11-17 17:52:19 · answer #10 · answered by Anonymous · 0⤊ 0⤋