how do i find the real or imaginary # to 125x^3-27=0

Question

how do i find the real or imaginary number to 125x^3-27=0

Answer 1

125x^3-27=0

this is the difference of two cubes

(5x)^3-(3^3)=0

a^3-b^3=(a-b)(a^2+ab+b^2)

(5x)^3-(3^3)=(5x-3)(25x^2+15x+9)=0

5x-3=0, 5x=3, x=3/5

(25x^2+15x+9)=0

using the quadratic equation

[-15+sqrt(225-900)]/50

[-15+sqrt(-675)]/50

[-15+sqrt(-3*225)]/50

[-15+15sqrt(-3)]/50

[-15+15i sqrt(3)]/50

[-3+3i sqrt(3)]/10

x=3/5, x=[-3+3i sqrt(3)]/10, and x=[-3-3i sqrt(3)]/10

Answer 2

125x^3-27 = 0

Factor the left side as a difference of two cubes:

a^3 - b^3 = (a-b)(a^2 +ab + b^2)

In the problem:

a = 5x and b = 3, since (5x)^3 = 125x^3 and 3^3 = 27

(5x-3)(25x^2 + 15x + 9) = 0

As William pointed out, setting 5x-3 = 0 gives you x = 3/5.


To get any other possible roots, you can do the following:

25x^2 + 15x + 9 = 0

-15 +- sqrt(225 - 4(25)(9))
x = ---------------------------------
50

x = -15 +- sqrt(225 - 900)
--------------------
50

x = -15 +- sqrt(-675)
-------------------
50

x = -15 +- sqrt( (-1)(25)(9)(3) )
--------------------------------
50

x = -15 +- 15i sqrt(3)
-------------------
50

x = -3 +- 3i √3
------------
10

Don't forget the 3/5:

x = {3/5, -3/10 + (3i√3)/10, -3/10 - (3i√3)/10}

Answer 3

Dear Micah,

 

Rewrite the equation as

 

125x³ = 27

 

Divide by 125

 

x³ = 27/125

 

x = (27/125)^1/3

 

The author of this problem did you a huge favor.  The numerator, 27, is 3³ the denominator, 125 is 5³.

 

So the cube root of 27/125 is 3/5