Find a zero of a function: $f(x)= x+\sqrt{1-x}$

Question

Find a zero of a function: $$f(x)= x+\sqrt{1-x}$$

What I did: $$x+\sqrt{1-x}=0$$

$$\sqrt{1-x}=-x$$

$$1-x=x^2$$

$$0=x^2+x-1$$

$$x_{1}=\frac{-1+\sqrt{5}}{2} $$ $$x_{2}=\frac{-1-\sqrt{5}}{2} $$

But in solutions says there's only one zero of this function: $$x_{2}=-\frac{1}{2}-\frac{\sqrt{5}}{2} $$

And I can't figure out why? The domain is $(-\infty,1]$ and both $x_{1}$ and $x_{2}$ "lie inside" that domain.

It can be even seen from the graph that this function has only one zero $(-\frac{1}{2}-\frac{\sqrt{5}}{2},0)$:

Answer 1

The problem is in the line $\sqrt{1-x}=-x$. On the left hand side you have always something positive, but this has to be equal with $-x$, so $x$ must be negative.

Answer 2

Anytime you square during your steps, you are liable to gain extraneous solutions, so you have to go back and make sure that the solutions work.

Answer 3

You can see that in the $f(x)$ values for $x \ge 1$ are not defined .