How do you solve and check for extraneous solutions in $sqrt(x+3) =x-3$ ?

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Answer 1 · 2015-07-31T21:52:25

Square both sides, solve the resulting quadratic equation and discard the extraneous solution to find: $x=6$

First note that we require $x >= 3$ since the left hand side is non-negative.

Now square both sides to get:
$x+3 = x^2-6x+9$

Subtract $x+3$ from both sides to get:

$0 = x^2-7x+6 = (x-1)(x-6)$

This has two solutions: $x=1$ and $x=6$ , but $x=1$ is not a solution of the original equation since we require $x >= 3$ .

So the only solution to the original equation is $x=6$ .

How do you solve and check for extraneous solutions in sqrt(x+3) =x-3sqrt(x+3) =x-3?