what are real roots?

Question

if the cubic eqn is : 2x^3 + 6x^2 - 4.5x - 13.5

what are the roots and which ones are real.
if changes had to be made to this equation in order to find out:
1) 2 real roots
2) 1 real roots
3) 1 complex root

what would the equations be??

Answer 1

2x^3 + 6x^2 - 4.5x - 13.5 =
2x^2(x + 3) - 4.5(x + 3) =
(2x^2 - 4.5)(x + 3) =
(x√2 + √4.5)(x√2 - √4.5)(x + 3)
x = -1.5, 1.5, -3
4x^3 + 12x^2 - 9x - 27 =
4x^2(x + 3) - 9(x + 3) =
(4x^2 - 9)(x + 3) =
(2x + 3)(2x - 3)(x + 3)
These are ways to change the equation to reduce the number of real roots to 2 and still have a cubic:
(x + 3)(2x + 3)^2
(x + 3)(2x - 3)^2
(2x + 3)(x + 3)^2
(2x - 3)(x + 3)^2
For one real root,
(4x^2 + 9)(x + 3)
(x + 3)^3
(2x + 3)^3
(2x - 3)^3
A polynomial can only have 1 complex root if it has imaginary coefficients.
(2x + 3)(2x - 3)(x + i3) is one of 6 possibilities with these factors.

Answer 2

2x^3 + 6x^2 - 4.5x - 13.5
its not an equation but just an expression.
2x^3 + 6x^2 - 4.5x - 13.5 = 0
is an equation.

we have

2x^2(x + 3) - 4.5(x + 3) = 0

=> 2x^2 - 4.5 = 0
or x + 3 = 0 => x = -3 (first real root)

=> 2x^2 = 4.5

=> x = +/-sqrt(4.5/2)

I dont think there are any complex root here.

Complex roots are roots that are comlex.

They must be in the form of a+ib a, and b must be real numbers.

You got me at the conversion. Sorry.

Peace out.

Answer 3

If you factor it...
2x^2(x + 3) - 4.5(x+3) ---> (2x^2 - 4.5) (x+3)

So you get -3, sqrt(4.5/2) so you get two real roots since you aren't taking the sqrt of any negative numbers.