Question

What are the real or imaginary solutions of the polynomial equation
64x^3 + 27 = 0
1. 3/4, -3/4
2.-3/4, 3± 3i√3/8
3. no solution
4. 3/4, 12±12i√3/32

Answer 1

Have you considered factoring the Difference of Cubes?

Answer 2

um, I'm not sure how to do that

Answer 3

we can rewrite the left side of the equation, like below:
\[\Large 64{x^3} + 27 = \left( {4x + 3} \right) \cdot \left( {16{x^2} - 12x + 9} \right)\]

Answer 4

I see, after that what do I do?

Answer 5

so, we can rewrite that equation as below:
\[\Large \left( {4x + 3} \right) \cdot \left( {16{x^2} - 12x + 9} \right) = 0\]
which is equivalent to these equations:
\[\Large \begin{gathered}
4x + 3 = 0 \hfill \\
16{x^2} - 12x + 9 = 0 \hfill \\
\end{gathered} \]
now, the first equation gives the real solution \(x=-3/4\), so the imaginary or complex solutions are given, solving the second equation:
\[\Large 16{x^2} - 12x + 9 = 0\]

Answer 6

So it would be #1?

Answer 7

I think no, since option #1 doesn't contain imaginary or complex numbers

Answer 8

oops duh. So it's 3 or 4

Answer 9

x=-3/4 is the real solution. Now we have to solve this equation:
\[\Large 16{x^2} - 12x + 9 = 0\]
using the standard formula:
\[\Large x = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
where \(a=16,b=-12,c=9\). Please try to solve such quadratic equation

Answer 10

it's 4