HaoQIN That is technically correct, though quite high-level and I don’t recommend using it. However, this is a potential solution. To anyone who wants to read this, Mr. Qin’s solution is as follows:
As f(x) = |x + 1| is a piecewise function defined as when x >= -1, f(x) = x + 1 and when x < -1, f(x) = -(x + 1), we can take the square of both sides to eliminate the positive-negative enigma, and as the solution to a quadratic equation is \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}, the sum of two roots would be \dfrac{-b}{a}, and plugging in a = 1 and b = 2 due to the expansion of the equation leading to x^2 + 2x + 1, the sum of two roots are -2.
For another case, we can also use the graph of an absolute function (you can search it up). As the graph is shifted one unit to the left, all values decrease by -1 in the function, and as the sum of the two solutions to |x| = k was 0, now for |x + 1| = k it would be -2, except for when |x + 1| = 0, where it would be -1.
