Question

Find all real solutions and show work: 2x^3-3x^2+18x-27=0

Answer 1

Given:

The given equation is

$2x^3-3x^2+18x-27=0$

To find:

All the real solutions.

Solution:

We have,

$2x^3-3x^2+18x-27=0$

It can be written as

$x^2(2x-3)+9(2x-3)=0$

$(2x-3)(x^2+9)=0$

Using zero product property, we get

$(2x-3)=0$ and $(x^2+9)=0$

$2x=3$ and $x^2=-9$

$x=\dfrac{3}{2}$ and $x=\pm\sqrt{-9}$

We know that $x=\pm\sqrt{-9}$ is an imaginary number because there is a negative sign under the square root.

Therefore, $x=\dfrac{3}{2}$ is the only real solution of the given equation.