Is there a difference between $\sqrt{x+2}+x=0$ and $x^2-x-2=0$

Question

Is there a difference between $\sqrt{x+2}+x=0$ and $x^2-x-2=0$

Solutions are $x=2$ or $x=-1$.

But $x=2$ does not satisfy $\sqrt{x+2}+x=0$.Because $\sqrt{4}+2 \neq0$

So does it mean that they are different ? Why ?

Answer 1

$$\sqrt{x+2}+x=0$$

$$\sqrt{x+2}=-x$$

But because $\sqrt{x}$ is taken as the principal root we need $-2 \leq x \leq 0$. That is the difference. Squaring introduces extraneous solutions.

Answer 2

Yes. By convention $\sqrt{x+2}$ is non-negative.

You can multiply the first by $\sqrt{x+2}-x$ to get $x+2-x^2=0$. You introduced the solution $x=2$ when you assumed $\sqrt{x+2}\sqrt{x+2}=x+2$. This is not true, since $x+2$ can be negative.

Note that this does not violate the fundamental theorem of algebra since you did not begin with a polynomial.

Answer 3

Ofcourse there lies a difference,

Starting with $ x^2-x-2=0 \implies x^2=x+2 $ , Now follow from here that ,

$ x = \pm \sqrt{x+2} $ , When you take the negative value you get the above result as the solutions are $ x=2 $ or $x=-1 $ , so the negative value satisfies when you have a equation created with the negative sign taken.

In particular , $ (x+\sqrt{x+2})(x-\sqrt{x+2}) = x^2-x-2=0 $ here, so they doesn't represent the same equation.