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

To find the number of real solutions for the equation: \(x|x+3| + |x-1| - 2 = 0\), we need to analyze the behavior of the expression based on the values of \(x\) which affect the absolute values.

The critical points for changes in the expression are \(x = -3\) and \(x = 1\). These are the points where the expressions within the absolute values change signs. We will evaluate the function in the following intervals:

1. \(x < -3\)
2. \(-3 \leq x < 1\)
3. \(x \geq 1\)

1. When \(x < -3\):

• \(|x+3| = -(x+3)\) because \(x+3\) is negative.
• \(|x-1| = -(x-1)\) because \(x-1\) is negative.
• The equation becomes \(x(-(x+3)) - (x-1) - 2 = 0\).
• Simplifying gives: \(-x^2 - 3x - x + 1 - 2 = 0 \Rightarrow -x^2 - 4x - 1 = 0\).
• This quadratic does not have real roots, as its discriminant is negative.

2. When \(-3 \leq x < 1\):

• \(|x+3| = x+3\) because \(x+3\) is non-negative.
• \(|x-1| = -(x-1)\) because \(x-1\) is negative.
• The equation becomes \(x(x+3) - (x-1) - 2 = 0\).
• Simplifying gives: \(x^2 + 3x - x + 1 - 2 = 0 \Rightarrow x^2 + 2x - 1 = 0\).
• Using the quadratic formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = 2\), \(c = -1\):
• The roots are \(x = \frac{-2 \pm 2\sqrt{2}}{2}\), resulting in \(x = -1 \pm \sqrt{2}\) (within bounds).

3. When \(x \geq 1\):

• \(|x+3| = x+3\) because \(x+3\) is non-negative.
• \(|x-1| = x-1\) because \(x-1\) is non-negative.
• The equation becomes \(x(x+3) + (x-1) - 2 = 0\).
• Simplifying gives: \(x^2 + 3x + x - 1 - 2 = 0 \Rightarrow x^2 + 4x - 3 = 0\).
• Using the quadratic formula, the roots are \(x = \frac{-4 \pm \sqrt{16 + 12}}{2} = \frac{-4 \pm \sqrt{28}}{2} = \frac{-4 \pm 2\sqrt{7}}{2}\), resulting in \(x = -2 \pm \sqrt{7}\).
• The positive root is valid as it falls in our defined half-open interval.

Upon analyzing these intervals, we find 3 real solutions:

1. \(x = -1 - \sqrt{2}\)
2. \(x = -1 + \sqrt{2}\)
3. \(x = -2 + \sqrt{7}\)

Thus, the number of real solutions is 3.