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

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2 Answers

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}
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 3the denominator, 125 is 53.
 
So the cube root of 27/125 is 3/5