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

Jun 27, 2015

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

${a}^{2} = - 2 a$

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:

${a}^{2} = - 2 a$

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:

$\sqrt{- 2 a} = \sqrt{0} = 0 = a$

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

When $a = - 2$ we have:

$\sqrt{- 2 a} = \sqrt{- 2 \cdot - 2} = \sqrt{4} = 2 \ne - 2 = a$

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