Find the roots of the equation

Question

128-2x³=0

Answer 1

There are actually three roots for this equation, since you have a cubic (degree 3) polynomial.

 

You have 128 - 2x^3 = 0, or x^3 - 64 = 0. This can be factored as (x-4)(x^2 + 4x + 16) = 0.

 

So you have x-4 = 0 and x^2 + 4x + 16 = 0. For x-4=0, it is fairly obvious that x = 4 is a solution. For x^2 + 4x + 16, you may use the quadratic formula to find the two other roots/solutions.

Answer 2

Hi Nina,

 

128 - 2x^3 = 0

 

2x^3 = 128

 

x^3 = 64

 

x = 64 ^ (1/3) = the cube root of 64 = 4

 

since 4 x 4 x 4 =64

 

Sincerely,

 

Ralph K.