20x^3=125x
....so far i got
20x^3 - 125x=0
5x(4x^2 - 25)=0
where do i go from here?
2007-08-23 06:55:44 · 6 answers · asked by hidden_within_a_nightmare 3 in Science & Mathematics ➔ Mathematics
I'd do it this way:
20x^3 - 125x = 0
5x(4x^2 - 25) = 0
4x^2 - 25 = 0
(2x -5)(2x + 5) = 0
x = 5/2, -5/2, and 0
You were heading in the right direction, just needed to factor the x^2 term.
2007-08-23 07:01:59 · answer #1 · answered by Jeremiah F 3 · 0⤊ 0⤋
20x^3 - 125x=0
5x(4x^2 - 25)=0
5x(2x+5)(2x-5)=0
5x = 0 OR 2x+5 = 0 OR 2x-5 = 0
=> x=0 OR x = -5/2 OR x = +5/2
2007-08-23 14:01:09 · answer #2 · answered by harry m 6 · 0⤊ 0⤋
20 x³ - 125 x = 0
( 5 x )(4 x ² - 25) = 0
x = 0 , x ² = 25 / 4
x = 0 , x = ± ( 5 / 2 )
2007-08-27 11:55:21 · answer #3 · answered by Como 7 · 0⤊ 0⤋
There can be two results,
5x = 0 ---> x = 0
or
4x^2 - 25 = 0 ----> x^2 = 25/4 ---> x = +/- 5/2
2007-08-23 14:01:38 · answer #4 · answered by Anonymous · 0⤊ 0⤋
Now Either 5 x = 0 --------> x = 0
Or ( 4 x ^ 2 - 25) = 0
--> ( 2x - 5 ) ( 2x + 5) = 0
Either 2 x = 5 ---> x = 5 / 2 -----> x = 2.5
Or x = - 5 / 2 ----> x = - 2.5
2007-08-23 14:01:36 · answer #5 · answered by H . B . K 2 · 0⤊ 0⤋
5x(4x^2-25)=0
so either 5x=0 or 4x^-25=0
or x=0 or 4x^2-5^2=0
or x=0 or (2x+5)=0 or 2x-5=0
or x=0 or x=-5/2 or x=5/2 ans
2007-08-23 14:03:30 · answer #6 · answered by MAHAANIM07 4 · 0⤊ 0⤋