How do i solve the equation by factoring?
20x^3-20x=9x^2
^ means to the # power
thank you
2007-02-23 07:24:52 · 8 answers · asked by abc 1 in Science & Mathematics ➔ Mathematics
rewrite to say:
20x^3-9x^2-20x=0
x(20x^2-9x-20)=0
x(5x+4)(4x-5)=0
x=0,-4/5,5/4
2007-02-23 07:35:19 · answer #1 · answered by Anonymous · 0⤊ 0⤋
You need to put all the "x" terms on one side of the equal sign. So subtract 9x^2 from both sides to get:
20x^3-9x^2-20x=0
Now you have to factor and solve for x.
[Hint: first factor out x to get:: x(20x^2-9x-20)=0]
FOR THE RECORD... I'm not trying to do his homework for him!!! I'm trying to show him how to set up the problem
2007-02-23 15:33:02 · answer #2 · answered by pandorius 2 · 0⤊ 1⤋
steps
1.) bring 9x^2 to the left side . so now it will be 20x^3-9x^2-20x=0
2.) divide x on both sides . then it will be 20x^2-9x-20=0(ax^2+bx=c)
3.) (x-25) (x+16)=0
4.) x could be 25 or -16
2007-02-23 15:38:10 · answer #3 · answered by nirmala 1 · 0⤊ 0⤋
20x^3-20x=9x^2
note that x = 0 is a solution
next divide the x away you will get
20x^2-20=9x
wich is same as
20x^2-20-9x=0
this can be solved with the abc formula :
x1 = [ 9 + sqrt (81 + 4*20*20) ] / 2*20
x2 = [ 9 - sqrt (81 + 4*20*20) ] / 2*20
sqrt (81 + 4*20*20) = 809
(and dont forgetr to mention that x3=0 is also a solution)
2007-02-23 15:34:12 · answer #4 · answered by gjmb1960 7 · 0⤊ 1⤋
20x^3-20x=9x^2
x(20x^2 -9x -20)
x(x - 5/4)(x + 4/5)
2007-02-23 15:35:48 · answer #5 · answered by Henry 4 · 0⤊ 0⤋
20x^3 - 20x = 9x^2
20x^3 - 9x^2 - 20x = 0
x(20x^2 - 9x - 20) = 0
x(5x + 4)(4x - 5) = 0
x= 0 or 5/4, or -4/5
2007-02-23 15:33:27 · answer #6 · answered by Michael C 3 · 0⤊ 0⤋
20x^3 - 20x = 9x^2
20x^3 - 9x^2 - 20x = 0
x(20x^2 - 9x - 20) = 0
x(5x + 4)(4x - 5) = 0
x = 0, (-4/5), or (5/4)
2007-02-23 15:41:57 · answer #7 · answered by Sherman81 6 · 0⤊ 0⤋
x (5x+4) (4x-5) = 0
2007-02-23 15:28:37 · answer #8 · answered by danrouthier 2 · 0⤊ 1⤋