Question

-19x^3-12x^2+16x=0
find the real solutions of the equation by graphing

Answer 1

like doing the graph of the function by derivating and finding critical points?

Answer 2

how do you figure out or find the solution such as
-1.29
-1.29,0.65
0,1.29,-0.65
0,-1.29,0.65

Answer 3

-19x^3-12x^2+16x=0
-19x^2-12x+16=0
use quadratic formula or factor out to find roots

Answer 4

a=-19 b=-12 c=16
\[x = \frac{ -b \pm \sqrt{b ^{2}-4ac} }{2a }\]

Answer 5

x=-12+-(-12)^2-4(-19)(16)/2(-19)

Answer 6

actually, it will begin by x=12+-, because -b= -(-12)=12

Answer 7

and theres a square root

Answer 8

12+-144-76(64)/-38

Answer 9

\[x=\frac{ 12\pm \sqrt{144+4\times16\times19} }{ -2\times19}\]

Answer 10

x=-1.29, x=0 and x=0.65

Answer 11

dont forget the x=0, because at the beginning, you factored by x and then assumed either x = 0 or the other multiple is 0

Answer 12

but how did you get -1.29 and 0.65

Answer 13

i just calculated the number under the square root. then, it says +-, so one time you solve the equation with a +, and then another time with a -. gives 2 answers, plus the x=0

Answer 14

okay thats what i didnt do thanks so much for the help

Answer 15

youre welcome