Question

what is the solution set for x in the equation below?
\\(\sqrt{x + 1} - 1 = x\\)

1. \\(\\{1\\}\\)
2. \\(\\{0\\}\\)
3. \\(\\{-1, 0\\}\\)
4. \\(\\{0, 1\\}\\)

Explanation:

Step1: Isolate the radical term

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

Step2: Square both sides

$(\sqrt{x+1})^2 = (x+1)^2$
$x+1 = x^2 +2x +1$

Step3: Simplify to quadratic equation

$x^2 +x =0$
$x(x+1)=0$

Step4: Solve for x and verify

Solutions are $x=0$ and $x=-1$.

1. For $x=0$: $\sqrt{0+1}-1 =1-1=0$, matches the right-hand side.
2. For $x=-1$: $\sqrt{-1+1}-1 =0-1=-1$, matches the right-hand side.

Answer:

$\{-1, 0\}$ (Option 3)