Real?

If we square both sides we get:

$X-1 = (X+1)^2$ we can expand $\displaystyle (X+1)^2 = X^2 + 2X + 1$ [ The Binomial Theorem ] and we get the equation: $X-1 = X^2 + 2X + 1$ Adding 1 to both sides would yield $X = X^2 + 2X + 2$ subtracting $X$ from both sides gives us: $0 = X^2 + X + 2$ and clearly this is a quadratic equation whose two roots are (approximately) $X_1=-0.5 + -1.3228756555322953i$ and $X_2 = -0.5 + 1.3228756555322953i$ which is clearly not real.

I derived the two roots from the equation:

$X = \frac{-1 \pm \overbrace{\sqrt{1^2 - 4(1)(2)}}^{NOT \ REAL}}{2(1)} = -0.5 + 1.3228756555322953i _\square$

This was real simple! Get it?

The square root of a real number is only real if the real number is positive. Since we're subtracting $1$ from $x$ under the square root, $x$ must be greater than or eaqual to $1$ for the square root to be real. For $y \geq 1, y \in \mathbb{R}$ , $\sqrt{y} \leq y$ , so $\sqrt{x-1} \leq x-1 < x+1$ , so the right side is always strictly greater than the left side and therefore the equation has no real solutions.