Question 172238
Solve 

{{{x^2 + sqrt(22x) = 0}}}
<pre><font size = 4 color = "indigo"><b>
Isolate the radical term by subtracting {{{x^2}}} from both sides:

{{{sqrt(22x) = -x^2}}}

Square both sides:

{{{(sqrt(22x))^2=(-x^2)^2}}}

{{{22x=x^4}}}

{{{22x-x^4=0}}}

{{{x(22-x^3)=0}}}

{{{matrix(4,4,  x=0, "", "", 22-x^3=0,
      "", "", "", -x^3=-22,
      "", "", "", x^3=22,
      "", "", "",  x=root(3,22) )}}}

We must always check even-root radical equations,
because there may be extraneous solutions,

Checking {{{x=0}}}

{{{x^2 + sqrt(22x) = 0}}}
{{{0^2 + sqrt(22(0)) = 0}}}
{{{0 + sqrt(0) = 0}}}
{{{0 + 0 = 0}}}
{{{0=0}}}
root(3,22)
Checking {{{x=root(3,22)}}}

{{{x^2 + sqrt(22x) = 0}}}
{{{(root(3,22))^2 + sqrt(22root(3,22)) = 0}}}
{{{22^(2/3) + sqrt(22^1*22^(1/3))=0}}}
{{{22^(2/3) + sqrt(22^(1+1/3))=0}}}
{{{22^(2/3) + sqrt(22^(3/3+1/3))=0}}}
{{{22^(2/3) + sqrt(22^(4/3))=0}}}
{{{22^(2/3) + (22^(4/3))^(1/2)=0}}}
{{{22^(2/3) + 22^((4/3)(1/2))=0}}}
{{{22^(2/3) + 22^(2/3)=0}}}
{{{2*22^(2/3)=0}}}

That is false so {{{x=root(3,22)}}} is not a
solution.

So {{{x=0}}} is the only solution.

-------------------

Here is how we could have solved this much easier:

{{{x^2 + sqrt(22x) = 0}}}

Isolate the radical term by subtracting {{{x^2}}} from both sides:

{{{sqrt(22x) = -x^2}}}

A square root radical always represents a non-negative number,
so the left side is non-negative.  The right side, however, will
always be negative unless {{{x=0}}}.  So 0 is the only possible
solution, and we check it as above and find it is a solution,
so the only solution is {{{x=0}}}.

Edwin</pre>