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
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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) (Show Source): You can put this solution on YOUR website!
The answer is
. 0 is a solution of the original equation. –2 is an extraneous solution.
If you square both sides, you get the equation
That equation has all the solutions that has,
and then some.
The solutions for are and .
When we try them on ,
for .
Substituting into the equation ,
that give which verifies ,
so is a solution of the original equation.
However, makes
--> --> ,
so substituting into the equation
gives , which is not true.
So is an extraneous solution.
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