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

1 Answer
Wataru
Sep 28, 2014

sqrt{x-3}-sqrt{x}=3√x−3−√x=3

by subtracting sqrt{x-3}√x−3,

Rightarrow -sqrt{x}=3-sqrt{x-3}⇒−√x=3−√x−3

by squaring both sides,

Rightarrow x=9-6sqrt{x-3}+x-3⇒x=9−6√x−3+x−3

by cancelling out xx's,

Rightarrow0=6-6sqrt{x-3}⇒0=6−6√x−3

by dividing by 6,

Rightarrow 0=1-sqrt{x-3}⇒0=1−√x−3

by adding sqrt{x-3}√x−3,

Rightarrow sqrt{x-3}=1⇒√x−3=1

by squaring both sides,

Rightarrow x-3=1⇒x−3=1

by adding 3,

Rightarrow x=4⇒x=4

Even though all implications above are valid, some of them are not reversible. As a result, x=4x=4 we found does not satisfy the original equation, which makes it extraneous.

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