How do i solve 4x^3-36x=0 ?????????? i've been practicing curve sketching and i need to solve this in order to find the critical points. my brain wont tell me what to do and i know i know how to do this! please help me, im so frustrated!
2006-11-29 15:08:17 · 12 answers · asked by Crazy 4 Calc 1 in Science & Mathematics ➔ Mathematics
Factorize!
4x (x^2-9) = 0
4x (x+3) (x-3)
your solution is
X = 0
x= -3
x= 3
2006-11-29 15:11:09 · answer #1 · answered by Anonymous · 0⤊ 0⤋
Let's start by factoring common terms from the expression:
4x^3 - 36x = 4x(x^2 - 9)
Now it rolls from here. The term in parentheses is the difference of two squares.
4x(x^2 - 9) = 4x(x + 3)(x - 3)
So then, if 4x(x + 3)(x - 3) = 0, then x = {0, +3, -3}
Cheers!
2006-11-29 23:10:50 · answer #2 · answered by hokiejthweatt 3 · 0⤊ 0⤋
this is a way of solving your problem:
4x^3-36x =0
<=> 4x*(x^2-9) =0
<=> 4x*(x+3)*(x-3)=0
<=> x=0 or x=3 or x=-3
2006-11-30 02:32:27 · answer #3 · answered by Huynh Dinh Tri 2 · 0⤊ 0⤋
First factor out an x to get x(4x^2- 36)=0, and then factor by difference of 2 squares. x(2x+6)(2x-6)=0, so the critical points would be 0, -3, 3 .
2006-11-29 23:11:39 · answer #4 · answered by Macho-man 3 · 0⤊ 0⤋
x=0, x=3, x= -3:
4x^3 - 36x = 4x(x^2 - 9) = 4x(x - 3)(x + 3) =0
So x=0, x=3, x= -3:
2006-11-29 23:10:39 · answer #5 · answered by can_t_get_enough 2 · 0⤊ 1⤋
Factorise the equation:
4x(x+3)(x-3)=0
So x = 0, -3, or 3
2006-11-29 23:11:54 · answer #6 · answered by martina_ie 3 · 0⤊ 0⤋
Then
4x^3 = 36 x
and
4x^2 = 36
x^2 = 36/4 = 9
x = 3 didn't even need a calculator for that one.
2006-11-29 23:11:44 · answer #7 · answered by Roadkill 6 · 0⤊ 2⤋
take x out as common
x(4x^2-36)=0
x(2x-6)(2x+6)=0
critical points x=0, x=-3, x=3
2006-11-29 23:12:01 · answer #8 · answered by lol 1 · 0⤊ 0⤋
4x^3-36x=0
4x(x^2-9)=0
x(x+3)(x-3)=0
x=0
x=3
x=-3
are the 3 roots.
2006-11-29 23:20:40 · answer #9 · answered by yupchagee 7 · 0⤊ 0⤋
4x*(x^2 -9) =0
x=0, -3, 3
2006-11-29 23:10:48 · answer #10 · answered by Anonymous · 0⤊ 0⤋