Question

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

Answer 1

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.

Explanation:

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.