Given:
The given equation is
[tex]2x^3-3x^2+18x-27=0[/tex]
To find:
All the real solutions.
Solution:
We have,
[tex]2x^3-3x^2+18x-27=0[/tex]
It can be written as
[tex]x^2(2x-3)+9(2x-3)=0[/tex]
[tex](2x-3)(x^2+9)=0[/tex]
Using zero product property, we get
[tex](2x-3)=0[/tex] and [tex](x^2+9)=0[/tex]
[tex]2x=3[/tex] and [tex]x^2=-9[/tex]
[tex]x=\dfrac{3}{2}[/tex] and [tex]x=\pm\sqrt{-9}[/tex]
We know that [tex]x=\pm\sqrt{-9}[/tex] is an imaginary number because there is a negative sign under the square root.
Therefore, [tex]x=\dfrac{3}{2}[/tex] is the only real solution of the given equation.
