Question

Find the number of real solutions of \[ x|x-3|+|x-1|+3=0 \]

• A. 1
• B. 2
• C. 3
• D. 4

Hint:

Always split absolute value equations into cases based on critical points where expressions inside absolute values change sign.

Answer 1

The Correct Option is A

Step 1: Consider the sign of $x$.
If $x \ge 0$, then $x|x-3|\ge 0$, $|x-1|\ge 0$, and hence: \[ x|x-3|+|x-1|+3>0 \] So, no solution exists for $x \ge 0$. 
Step 2: Consider $x<0$. 
For $x<0$: \[ |x-3|=3-x,\quad |x-1|=1-x \] Substitute in the equation: \[ x(3-x)+(1-x)+3=0 \] Step 3: Simplify the equation. 
\[ 3x-x^2+4-x=0 \] \[ -x^2+2x+4=0 \] \[ x^2-2x-4=0 \] Step 4: Solve the quadratic equation. 
\[ x=1\pm\sqrt{5} \] Since $x<0$, only $x=1-\sqrt{5}$ is valid. 
Step 5: Count the number of real solutions. 
There is only one real solution.