Question 155348This question is from textbook College Algebra
: Please help me solve this. I need to put it in standard form then solve for x using the quadratic equation which should give me a complex number as an answer. The problem is: . Thanks.
This question is from textbook College Algebra
Found 2 solutions by checkley77, nerdybill:
Answer by checkley77(12844)   (Show Source):
You can put this solution on YOUR website! Don't fully understand the need for the standard form & the use of the quadratic equation solution when there is a more direct approach as follows:
x^3-8=0
x^3=8
x=cubert8
x=2
Answer by nerdybill(7384)  (Show Source):
You can put this solution on YOUR website! Notice that your equation:
is a "difference of two cubes"
That is, both terms have cube roots:
the cube root of x^3 is x
the cube root of 8^3 is 2
.
This site describes these "special cases":
http://www.purplemath.com/modules/specfact2.htm
.
Bottom-line:
If you see a situation of a "difference of two cubes", you can factor thus:
a^3 – b^3 = (a – b)(a^2 + ab + b^2)
.
In our case:
a is x
b is 2
.
Therefore, we can rewrite:
as
.
Finally, to solve, we set each factor on the left to zero:
First term:
(x-2) = 0
x = 2 (Here's one "real" solution)
.
Second term:
(x^2 + 2x + 4)=0
Here's where we need to apply the "quadratic equation":
Note: you will find that there are no real solutions, only 2 imaginary ones:
| Solved by pluggable solver: SOLVE quadratic equation with variable |
Quadratic equation  (in our case  ) has the following solutons:
For these solutions to exist, the discriminant  should not be a negative number.
First, we need to compute the discriminant :  .
The discriminant -12 is less than zero. That means that there are no solutions among real numbers.
If you are a student of advanced school algebra and are aware about imaginary numbers, read on.
In the field of imaginary numbers, the square root of -12 is + or -  .
The solution is 
Here's your graph:
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