x^3=4x how do you solve this type of equation?

Hi,

For this question, first subtract 4x from both sides to get:

x^3 - 4x = 0

Now, factor out a common factor of 'x' to get:

x(x^2 - 4) = 0

Each factor equals zero so....

x = 0

AND

x^2 - 4 = 0

Now factor the difference of two squares:

(x-2)(x+2)=0

Set each factor equal to zero:

x-2=0

x=2

OR

x+2=0

x=-2

I hope that helps you out! Please let me know if you have any other questions!

x³ = 4x
↓
0 = x³ - 4x
0 = x (x²-4)
0 = x (x-2)(x+2)
↓
x = {-2,0,2)

First, notice that the equation is valid for x=0
0^3 = 4(0) = 0

Next, divide both sides by x so seek any other value except x=0 (because we would not be allowed to divide by zero).

x^3 / x = 4x / x
x^2 = 4
x = +/- 2

The equation is valid for x = 2
2^3 = 4(2) = 8

The equation is valid for x = -2
(-2)^3 = 4(-2) = -8

x^3 = 4x
x^3 - 4x = 0
x (x^2 - 4) = 0
x = 0 , x^2 - 4 = 0
x^2 = 4
x = + - 2
The answers are x = 0, + -2.

x ³ - 4x = 0
x ( x ² - 4 ) = 0
x ( x - 2 ) ( x + 2 ) = 0
x = 0 , x = 2 , x = - 2

x^3-4x=0

x(x^2-4)=0
x(x-2)(x+2)=0
x=0x=2x=-2

x^3 = 4x
x^3 - 4x = 0
x(x^2) - x(4) = 0
x(x^2 - 4) = 0
x(x^2 - 2^2) = 0
x(x + 2)(x - 2) = 0

x = 0

x + 2 = 0
x = -2

x - 2 = 0
x = 2

∴ x = -2, 0 or 2

x^3 - 4x = 0
x ( x^2 - 4 ) = 0
x ( x + 2 ) ( x - 2 ) = 0
x = 0 , 2 , -2 ..