# How do I find the extraneous solution of sqrt(x-3)-sqrt(x)=3?

Sep 28, 2014

$\sqrt{x - 3} - \sqrt{x} = 3$

by subtracting $\sqrt{x - 3}$,

$R i g h t a r r o w - \sqrt{x} = 3 - \sqrt{x - 3}$

by squaring both sides,

$R i g h t a r r o w x = 9 - 6 \sqrt{x - 3} + x - 3$

by cancelling out $x$'s,

$R i g h t a r r o w 0 = 6 - 6 \sqrt{x - 3}$

by dividing by 6,

$R i g h t a r r o w 0 = 1 - \sqrt{x - 3}$

by adding $\sqrt{x - 3}$,

$R i g h t a r r o w \sqrt{x - 3} = 1$

by squaring both sides,

$R i g h t a r r o w x - 3 = 1$

$R i g h t a r r o w x = 4$
Even though all implications above are valid, some of them are not reversible. As a result, $x = 4$ we found does not satisfy the original equation, which makes it extraneous.