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

To find the number of real solutions of the equation:

\[x|x-3| + |x-1| + 3 = 0\]

we will analyze the expression by splitting it into different cases based on the critical points where the absolute value expressions change.

Case 1: \( x < 1 \)

• Both expressions \( |x-3| \) and \( |x-1| \) are negative:
• \( x|x-3| = x(3-x) = 3x - x^2 \) and
• \( |x-1| = 1-x \).
• The equation becomes:
• Solving the quadratic \( -x^2 + 2x + 4 = 0 \), we calculate the discriminant:
• The discriminant is positive, indicating two real roots, but they must both be less than 1 in this case:
• Hence, only \( 1 - \sqrt{5} \) falls under \( x < 1 \).

Case 2: \( 1 \leq x < 3 \)

• Here, \( |x-3| = 3-x \) (negative) and \( |x-1| = x-1 \) (positive).
• The equation becomes:
• Solving \( -x^2 + 4x + 2 = 0 \), calculate the discriminant:
• The discriminant is positive, thus two real roots. We solve:
• Both solutions are in the interval \( 1 \leq x < 3 \), but on evaluating they are \( 2 \pm \sqrt{2} \).

Case 3: \( x \ge 3 \)

• Both expressions \( |x-3| \) and \( |x-1| \) are positive:
• \( x|x-3| = x(x-3) = x^2 - 3x \) and
• \( |x-1| = x-1 \).
• The equation becomes:
• Solving \( x^2 - 2x + 2 = 0 \):
• The discriminant is negative, indicating no real roots in this interval.

In conclusion, only the solution \( x = 2 - \sqrt{2} \) lies in the desired range from Case 2, thus there is only one real solution.