2007-11-09 20:56:48 · 6 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
x^2+7x+12=0
x^2+(3x+4x)+12=0
(x^2+3x)+(4x+12)=0 grouping
x(x+3)+4(x+3)=0 factoring
(x+4)(x+3)=0
Set each factor to zero.
x+4=0 and x+3=0
the value of x and y is
x= -4 and x = -3
2007-11-09 22:32:23 · answer #1 · answered by nildo 1 · 1⤊ 1⤋
x² + 7x + 12 = 0
Sum of the roots -b/a = -7/1 = -7
Product of the roots = ac = 1 * 12 = 12
Roots = (-b ± (âb²-4ac))/2a where a , b and c are coefficients of the binomial expression , ie: a = 1 b = 7c = 12
(-(7 ) ± â(7)² - 4(1)(12) )/ 2(1)
= (-7 ± (â49-48))/2
= (-7 ± â1)/2
= (-7 ± 1)/2
= (-7+1)/2 and
(-7 - 1)/2
= -6/2 and -8/2
The roots are -3 and -4
Sum of the roots = -7
Products of the root = 12.
2007-11-09 21:27:51 · answer #2 · answered by A Little Sarcasm Helps 5 · 0⤊ 1⤋
(x+3)(x+4)
x = -3, -4
2007-11-09 21:01:44 · answer #3 · answered by JP 6 · 0⤊ 0⤋
x^2+7x+12=0
x^2+(3x+4x)+12=0
(x^2+3x)+(4x+12)=0
x(x+3)+4(x+3)
(x+4)(x+3)
hope that helped you... :)
2007-11-09 21:02:29 · answer #4 · answered by Muslim Girl 2 · 0⤊ 0⤋
x= square root of -7/12
2007-11-09 21:03:02 · answer #5 · answered by Anonymous · 0⤊ 0⤋
(x + 4)(x + 3) = 0
x = - 4 , x 0 - 3
2007-11-09 22:29:38 · answer #6 · answered by Como 7 · 2⤊ 1⤋