Question

Pleeease help! I have no clue how to even attempt this and I really need to get it done. :'( Please show all steps! Will give medal to best answer!

What are the real or imaginary solutions of the polynomial equation?

27x^3+125=0

Answer 1

Oh they want imaginary roots too!

27x^3 + 125 = 0
Divide each term by 27
x^3 + 125/27 = 0
Both 125 and 27 are perfect cubes. 125 = 5 x 5 x 5 and 27 = 3 x 3 x 3
x^3 + (5/3)^3 = 0

There is a formula: a^3 + b^3 = (a + b)(a^2 - ab + b^2)

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

x + 5/3 = 0 gives x = -5/3 (This is one root. There are two more)

x^2 - 5/3x + 25/9 = 0 is a quadratic equation.

a = 1, b = -5/3 and c = 25/9

Can you find the two roots of the above quadratic equation using the standard formula?

Answer 2

So was the work in your last comment incorrect? I wrote it all down. I think I forgot what the standard formula is..

Answer 3

wait, is it ax^2+bx+c?

Answer 4

Since they want the imaginary roots as well, this is a better way to proceed.
\[\Large ax^2 + bx + c = 0\] is the standard quadratic equation.

The roots for that equation is given by the formula:
\[\Large \frac{ -b \pm \sqrt{b ^{2} - 4ac} }{ 2a } \]

Put a = 1, b = -5/3 and c = 25/9 in the above formula.

Answer 5

Okay hang on let me try.

Answer 6

Okay.. I plugged it all in but seeing the fractions in there it totally throws me off..

Answer 7

1x^2-5/3x+25/9

Answer 8

1x^2-5/3x+25/9 = 0
You can easily get rid of the fractions by multiplying throughout by 9
9x^2 - 15x + 25 = 0

Now a = 9, b = -15, c = 25

Plug it into the formula. You will get a negative value inside the square root. So the two solutions will be imaginary. Remember that the square root of -1 is the symbol "i".

Answer 9

The one real solution is: -5/3

The two imaginary solutions are: \[\Large \frac{ 5 }{ 6 }(1 + i \sqrt{3})\]and\[\Large \frac{ 5 }{ 6 }(1 - i \sqrt{3})\]where \[\Large i = \sqrt{-1}\]