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Solve the equation. Identify any extraneous solutions.
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a = 0
There is an extraneous solution that emerges from the standard procedure of initially squaring both sides.
Do you know how to square both sides?
0 and –2 are solutions of the original equation. or 0 is a solution of the original equation. –2 is an extraneous solution.?
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yes
yes what?
??????????????
a = 0 a = √2 √-a -2a = a^2 -2a - a^2 = 0 -a(a + 2) = 0 So then you have: a = 0 OR a + 2 = 0 (subtract 2 from both sides) a = 0 OR a = -2
0 is the solution. -2 is the extraneous solution
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Did I do my work correctly @BangkokGarrett? Just curious?
Ok thanks guys:) im closing this now but im opening another.
Okay:) and you are welcome!
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