Solve the equation
2006-07-23 15:23:33 · 10 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
The given equation is
6x^2+5x+1 = 0
factorise this, 6x^2+3x+2x+1 = 0
3x(2x+1) + (2x+1) = 0
(2x+1)(3x+1) = 0
2x+1 = 0 and 3x + 1 = 0
hence, x = -(1/2) or x = -(1/3)
2006-07-23 16:59:37 · answer #1 · answered by Subhash G 2 · 7⤊ 0⤋
Use quadratic formula:
If ax^2 + bx + c = 0 , then x = [-b +/- sqrt(b^2 - 4ac)] / 2a
here a = 6, b = 5 and c = 1
x = [-5 +/- sqrt (5^2 - 4(6)(1))] / 2*6
= [-5 +/- sqrt (25 - 24)] / 12
= [-5 +/- sqrt (1)] / 12
= [-5 +/- 1] / 12
= -6/12 or -4/12
x = (-1/2) or (-1/3)
2006-07-23 23:07:30 · answer #2 · answered by Anonymous · 0⤊ 0⤋
6x^2 + 5x +1 = 0
(3x + 1)(2x + 1) = 0
3x + 1 = 0
3x = -1
x = -1/3
2x + 1 = 0
2x = -1
x = -1/2
* x = -1/2 or -1/3
2006-07-24 00:18:15 · answer #3 · answered by ping_er_18 1 · 0⤊ 0⤋
(x+a)(x+b)=x^2+x(a+b)+ab
ur eqn 6x^2+5x+1=0
.(3x+1)(2x+1)=6x^2+5x+1
(3x+1)=0 ;(2x+1)=0
3x=-1 ;2x=-1
x=-1/3 ; x=-1/2
x={-1/3,-1/2}
2006-07-24 00:04:47 · answer #4 · answered by corrona 3 · 0⤊ 0⤋
factorize
(3x+1)(2x+1)=0
If those two function's multiplication is to be 0, then either one can be 0
(3x+1)=0 , (2x+1)=0
x=-1/3 , x=-1/2
2006-07-23 22:37:26 · answer #5 · answered by sumone^^ 3 · 0⤊ 0⤋
-(1/3) or -(1/2)
2006-07-23 22:26:25 · answer #6 · answered by zzzzzzzzz 3 · 0⤊ 0⤋
6x^2 + 5x + 1 = 0
x = (-b ± sqrt(b^2 - 4ac))/2a
x = (-5 ± sqrt((5)^2 - 4(6)(1)))/(2(6))
x = (-5 ± sqrt(25 - 24))/12
x = (-5 ± sqrt(1))/12
x = (-5 ± 1)/12
X = (-6/12) or (-4/12)
x = (-1/2) or (-1/3)
2006-07-23 22:43:47 · answer #7 · answered by Sherman81 6 · 0⤊ 0⤋
(3x+1)(2x+1)=0
3x+1=0 2x+1=0
x=-1/3 x=-1/2
x={-1/2,-1/3}
2006-07-23 22:28:26 · answer #8 · answered by jogimo2 3 · 0⤊ 0⤋
factor or use the quadratic equation
head start, consider this factoring
(3x+1)(2x+1)
2006-07-23 22:30:17 · answer #9 · answered by enginerd 6 · 0⤊ 0⤋
=2 maybe?
2006-07-23 22:28:02 · answer #10 · answered by xx_dragonz_xx 3 · 0⤊ 0⤋