Question

How do I correct my mistake:

Answer 1

\[x = \sqrt{x + 2}\]
\[x^2 = x+2\]
\[x^2 - x - 2 = 0\]
\[(x+1)(x-2) = 0\]
\[x = -1 ,2\]

Answer 2

This isn't solving the equation, this is a mathematical proof.

Ex: Since x=-1 would make x = sqrt(x+2) false, then this proof is a failure

Answer 3

Not sure what mistake you're referring to

Since we have a square root, we need to plug in our answer choices to make sure that it is indeed a solution. And sqrt(1) \(\neq\) -1 so we know that x \(\neq -1\) and that the only solution is x = 2

Answer 4

Yes but the problem is, I can't take out x=-1

Answer 5

It's a proof logic:

If A then B
If A and B then C

or

If and only if A then B
B is false
Then A has to be false

Answer 6

But rational functions have 2 solutions

Answer 7

\[x = \sqrt{x + 2}\] and \[x^2 = x+2\] is not bi conditional basically, I got to figure out how to make it biconditional

Answer 8

Yeah dude, but there's an extraneous solution

Answer 9

\(-1=\pm \sqrt{-1+2}\)

Answer 10

Took me a while to type this

On mobile ;(

Answer 11

I can't change the original equation, I need to change a step in the proof instead

Answer 12

This is propositional logic, not solving an equation. I'm trying to prove something not solve it.

Answer 13

I see, mhm
maybe @Vocaloid can give some insight

Answer 14

Update: I got the answer. It's a paragraph btw so I wont bother typing it out