I would factor it.
(3x+2)(x-4)=0
x= -2/3, or x=4
Determine the zeros of the function 3x^2 -8=10x. I know I rewrite it as 3x^2 -10x-8 = 0
But then what do I do-I'm totally confused-Answers are
{-2,4}
{2,4}
{-2/3,4}
{2/3,4}
Please direct me or show the steps, just don't give me the answer because I still won't get it.I think it is -2/3,4 but I'm not sure-even if that is correct, please show me the steps.
Thank you
2 answers
3x^2 - 10x - 8 = 0,
Factor using the A*C Method:
A * C = 3 * -8 = -24 = 2 * -12,
3x^2 + (2x - 12x) - 8 = 0.
Group the 4 terms into factorable pairs:
(3x^2 - 12x) + (2x - 8) = 0,
3x(x - 4) + 2(x - 4) = 0,
Factor (x - 4) from each term:
(x - 4) (3x + 2) = 0,
x - 4 = 0,
x = 4.
3x + 2 = 0,
3x = -2,
x = -2/3.
Solution set: x =-2/3, and x = 4.
Factor using the A*C Method:
A * C = 3 * -8 = -24 = 2 * -12,
3x^2 + (2x - 12x) - 8 = 0.
Group the 4 terms into factorable pairs:
(3x^2 - 12x) + (2x - 8) = 0,
3x(x - 4) + 2(x - 4) = 0,
Factor (x - 4) from each term:
(x - 4) (3x + 2) = 0,
x - 4 = 0,
x = 4.
3x + 2 = 0,
3x = -2,
x = -2/3.
Solution set: x =-2/3, and x = 4.
1 answer