Question 840889
{{{x^3=216}}}-->{{{x^3-216=0}}}-->{{{x^3-6^3=0}}}-->{{{(x-6)(x^2+6x+6^2)=0}}}-->{{{(x-6)(x^2+6x+36)=0}}}
{{{x=6}}} , which makes {{{(x-6)=0}}} is the real solution;
{{{x^2+6x+36=0}}} , which has no real solution yields two more imaginary ones.
 
{{{6x=9+x^2}}}-->{{{x^2-6x+9=0}}}-->{{{(x-3)^2=0}}} has {{{x=3}}} as a double real solution (no imaginary ones).
 
{{{x^2+3x+3=0}}} has no real solutions, just imaginary ones.
You would know that by calculating the discriminant.
A quadratic equation, that can be written as
{{{ax^2+bx+c=0}}} , has solutions that can be calculated using the quadratic formula:
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
The expression {{{b^2-4*a*c}}} is called the discriminant, and its sign will tell you about the solutions. A negative discriminant means no real solutions. A positive one means two (different) real solutions. If the discriminant is zero, there is just one (double) real solution (as happened for {{{x^2-6x+9=0}}} above).
In this case {{{a=1}}} , {{{b=3}}} , and {{{c=3}}} , and
{{{b^2-4*a*c=3^2*4*1*3=9-12=-3<0}}} , so no real solutions.
 
{{{x^3+125=0}}}-->{{{x^3+5^3=0}}}-->{{{(x+5)(x^2-5x+5^2)=0}}}
{{{x=-5}}} is the real solution; {{{x^2-5x+25=0}}} , which has no real solution yields two more imaginary ones.
 
{{{x^4-24x^2+100=0}}} is more complicated.
You could use the change of variable {{{y=x^2}}} to write the equation as
{{{y^2-24y+100=0}}} , and find that the solutions for {{{y}}} .
You could solve for {{{y}}} by completing the square, or by using the quadratic formula.
Either way, you will find
{{{y=12 +- sqrt(44)}}} (or an equivalent expression).
Since both {{{y}}} solutions are positive, you will have four real solutions to
{{{x^2=12 +- sqrt(44)}}} , and those are the solutions of
{{{x^4-24x^2+100=0}}} .

so there are two real solutions.