SOLUTION: Hi! On the following problem I keep getting the incorrect answer. I think I'm close to the right answer just making a small error somewhere. Can you help me? Solve the equati

Question 196635: Hi!
On the following problem I keep getting the incorrect answer. I think I'm close to the right answer just making a small error somewhere. Can you help me?
Solve the equation for the variable x. The constant a represents a positive real number.
a^3x^3 + 64 = 0
Found 2 solutions by Alan3354, jim_thompson5910:
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
The constant a represents a positive real number.
a^3x^3 + 64 = 0
(ax+4)*(a^2x^2 - 4ax + 16) = 0
x = -4/a
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Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)

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 -48 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 -48 is + or - .

The solution is , or
Here's your graph:

----------------
Reducing gives:
2 � isqrt(12)
Then taking the "a" into account:
x = (2 � 2i*sqrt(3))/a

Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
Start with the given equation.

Rewrite as

Rewrite as

Factor the left side using the sum of cubes formula

Recall, the sum of cubes formula is

Multiply and simplify

or Use the zero product property to break up the factors

-----------------------------------------------------

Let's solve the first equation:

Start with the first equation.

Subtract 4 from both sides.

Divide both sides by "a" to isolate "x".

So the first solution is

-----------------------------------------------------

Now let's solve the second equation:

Start with the second equation.

Notice we have a quadratic in the form of where , , and

Let's use the quadratic formula to solve for "x"

Start with the quadratic formula

Plug in , , and

Negate -4a to get 4a

Square -4a to get

Multiply

Combine like terms.

Factor -48 into

Break up the square root.

Simplify the square roots and replace with "i".

Note:

Rearrange the terms.

or Break up the "plus/minus"

or Reduce

So the next two solutions are or

================================================

Answer:

So the three solutions are , or
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