Question 705205
The answer is
{{{highlight(C)}}}. 0 is a solution of the original equation. –2 is an extraneous solution.
 
{{{a=sqrt(-2a)}}}
If you square both sides, you get the equation
{{{a^2=-2a}}}
That equation has all the solutions that {{{a=sqrt(-2a)}}} has,
and then some.
The solutions for {{{a^2=-2a}}} are {{{a=0}}} and {{{a=-2}}}.
 
When we try them on {{{a=sqrt(-2a)}}},
for {{{a=0}}} {{{sqrt(-2a)=0}}}.
Substituting into the equation {{{a=sqrt(-2a)}}},
that give {{{0=0}}} which verifies {{{a=sqrt(-2a)}}},
so {{{a=0}}} is a solution of the original equation.
 
However, {{{a=-2}}} makes
{{{sqrt(-2a)=sqrt(-2(-2))}}} --> {{{sqrt(-2a)=sqrt(4)}}} --> {{{sqrt(-2a)=2}}},
so substituting {{{a=-2}}} into the equation {{{a=sqrt(-2a)}}}
gives {{{-2=2}}}, which is not true.
So {{{a=-2}}} is an extraneous solution.