How do you solve the equation and identify any extraneous solutions for #a=sqrt(-2a)#?

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Answer 1 · 2015-06-27T15:29:06

First square both sides (which may add extraneous solutions) to get:

#a^2 = -2a#

One solution of this is #a=0#. The other solution #a=-2# is extraneous.

If we square both sides of the original equation then we get:

#a^2 = -2a#

This must hold in order that #a# satisfy the original equation, but it is not a sufficient condition, as we shall see.

This new equation has two solutions:

#a=0# and #a=-2#

When #a=0# we have:

#sqrt(-2a) = sqrt(0) = 0 = a#

So #a=0# is a solution of the original equation.

When #a=-2# we have:

#sqrt(-2a) = sqrt(-2*-2) = sqrt(4) = 2 != -2 = a#

So #a=-2# is not a solution of the original equation.