SOLUTION: Solve the equation. Identify any extraneous solutions. a=√(-2a) A. –2 is a solution of the original equation. 0 is an extraneous solution. B. 0 and –2 are solutions of

Algebra ->  Radicals -> SOLUTION: Solve the equation. Identify any extraneous solutions. a=√(-2a) A. –2 is a solution of the original equation. 0 is an extraneous solution. B. 0 and –2 are solutions of      Log On


   

Question 705205: Solve the equation. Identify any extraneous solutions. a=√(-2a)
A. –2 is a solution of the original equation. 0 is an extraneous solution.
B. 0 and –2 are solutions of the original equation.
C. 0 is a solution of the original equation. –2 is an extraneous solution.
D. 2 is a solution of the original equation. 0 is an extraneous solution.
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
The answer is
highlight%28C%29. 0 is a solution of the original equation. –2 is an extraneous solution.

a=sqrt%28-2a%29
If you square both sides, you get the equation
a%5E2=-2a
That equation has all the solutions that a=sqrt%28-2a%29 has,
and then some.
The solutions for a%5E2=-2a are a=0 and a=-2.

When we try them on a=sqrt%28-2a%29,
for a=0 sqrt%28-2a%29=0.
Substituting into the equation a=sqrt%28-2a%29,
that give 0=0 which verifies a=sqrt%28-2a%29,
so a=0 is a solution of the original equation.

However, a=-2 makes
sqrt%28-2a%29=sqrt%28-2%28-2%29%29 --> sqrt%28-2a%29=sqrt%284%29 --> sqrt%28-2a%29=2,
so substituting a=-2 into the equation a=sqrt%28-2a%29
gives -2=2, which is not true.
So a=-2 is an extraneous solution.