I'm told to solve this equation 9x-4= -9x^2
I'm not too sure how to do it, I tried changing it to standard form but I was lost there since there is no common factor.
If anyone could do the steps for it, I'd appreciate it!
2007-12-10 12:33:35 · 5 answers · asked by Nick B 2 in Science & Mathematics ➔ Mathematics
Start by making it equal to zero. Then we'll factor and solve.
9x-4=-9x^2
9x-4+9x^2 =-9x^2 +9x^2
9x^2+9x-4=0 (Look familiar?)
To factor ax^2+bx+c=0 we need to find numbers that have a product of |ac| and sum to |b|. So
|ac|= 36 and |b|=9
Factor 36 (2,18),(3,12),(4,9) Use (3,12).
Since c is negative, we know that only one factor is negative. In this case, our factors are (-3, 12). Add these into our original equation as middle terms and factor by grouping.
9x^2+9x-4=0
9x^2-3x+12x-4=0
3x(3x-1)+4(3x-1)=0
(3x+4)(3x-1)=0
In order to solve for x from this point we take each factor and make it equal zero.
3x+4=0 and 3x-1=0
x=-4/3 and x=1/3
Email if you have questions.
2007-12-10 13:15:42 · answer #1 · answered by yukonruby 2 · 0⤊ 0⤋
9x^2 +9x - 4 = 0
(3x-1)(3x+4)=0
3x-1=0 x = 1/3 or 3x+4=0 x=-4/3
2007-12-10 12:38:48 · answer #2 · answered by norman 7 · 0⤊ 0⤋
9x^2 + 9x - 4 = 0
(3x + 4)(3x - 1) = 0
3x + 4 = 0 and 3x - 1 = 0
3x = -4 and 3x = 1
x = -4/3 and x = 1/3
2007-12-10 12:37:01 · answer #3 · answered by Ms. Exxclusive 5 · 0⤊ 0⤋
if you rearrange it it gives simple quadratic equation
9x - 4 = -9x^2
9x^2 + 9x - 4 = 0
9x^2 + 12x - 3x - 4 = 0
3x(3x + 4) - 1(3x + 4) = 0
(3x + 4)(3x - 1) = 0
3x = -4 or
3x = 1
x = -4/3
x = 1/3
2007-12-10 12:40:42 · answer #4 · answered by mohanrao d 7 · 0⤊ 0⤋
9x-4=-9x^2
9x^2+9x-4=0
so the factors are...
(3x+4)(3x-1)=0
x=-4/3 or x=1/3
2007-12-10 12:43:13 · answer #5 · answered by novia_elwe 2 · 0⤊ 0⤋