The number of real solutions of the equation: x|x+3| + |x-1| - 2… — Solution | JEE Mathematics

MCQMediumJEE 2026Quadratic Equations in Complex Numbers

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

Answer

Correct answer:C

Standard Method

Given: $x|x+3| + |x-1| - 2 = 0$

Find: The number of real solutions.

Absolute value expressions must be split at the sign-changing points.

The critical points are

x=-3,\; x=0,\; x=1

Case 1: $x \ge 1$

x(x+3) + (x-1) - 2 = 0

x^2 + 4x - 3 = 0

From the extracted solution, the valid root in this interval is $x=1$ and the other root is rejected.

Case 2: $0 \le x < 1$

x(x+3) + (1-x) - 2 = 0

x^2 + 2x -1 = 0

x = \frac{-2+\sqrt{8}}{2}

This value lies in $[0,1)$ , so it is valid.

Case 3: $-3 \le x < 0$

-x(x+3) + (1-x) - 2 = 0

-x^2 -4x -1 = 0

From the extracted solution, the valid root is $x=-1$ .

Therefore, the valid solutions are $x=1,\; -1,\; \frac{-2+\sqrt{8}}{2}$ .

So, the total number of real solutions is $3$ . Hence, the correct option is C.](streamdown:incomplete-link)

Splitting the equation only at $x=-3$ and $x=1$ is incorrect because the factor $x$ also changes sign at $x=0$ . You must consider all sign-changing points before forming cases.
Accepting roots without checking their interval is wrong because each reduced equation is valid only in its own case. After solving in an interval, verify that the root actually belongs to that interval.
Removing absolute values directly without sign analysis leads to incorrect equations. First decide the signs of $x+3$ and $x-1$ in each interval, then rewrite the expression accordingly.

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