How do you solve $sqrt(x+1) = x-1$ and find any extraneous solutions?

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Answer 1 · 2016-05-21T14:40:41

$x=3$
$x=0$

First, to remove the $sqrt$ , square both sides of the equation, giving:

$x+1=(x-1)^2$

Next, expand the equation out.

$x+1=x^2-2x+1$

Simplify the equation combining like terms.

$x^2-3x=0$

$x(x-3)=0$

Now, you can solve for $x$ :

$x=0$
$x=3$

However, if you solved it like this:

$x^2-3x=0$

$x^2=3x$

$x=3$

$x=0$ would be a missing solution, this would be an extraneous solution.

How do you solve sqrt(x+1) = x-1sqrt(x+1) = x-1 and find any extraneous solutions?