Question

The number of real solutions of the equation: $x|x+3| + |x-1| - 2 = 0$ is

• A. 5
• B. 4
• C. 3
• D. 2

Hint:

Always split absolute value equations at sign-changing points before solving.

Answer 1

The Correct Option is C

Step 1: Identifying critical points.
The expressions change sign at \[ x=-3,\; x=0,\; x=1 \] Step 2: Solving in different intervals.
Case 1: $x \ge 1$
\[ x(x+3) + (x-1) - 2 = 0 \] \[ x^2 + 4x - 3 = 0 \Rightarrow x= -2 \text{ (rejected)},\; x=1 \] Case 2: $0 \le x<1$
\[ x(x+3) + (1-x) - 2 = 0 \] \[ x^2 + 2x -1 = 0 \Rightarrow x = \frac{-2+\sqrt{8}}{2} \] Case 3: $-3 \le x<0$
\[ -x(x+3) + (1-x) - 2 = 0 \] \[ -x^2 -4x -1 = 0 \Rightarrow x=-1 \] Step 3: Counting valid solutions.
Valid solutions are \[ x = 1,\; -1,\; \frac{-2+\sqrt{8}}{2} \] Total number of real solutions $= 3$.