The number of the real solutions of the equation:
$x|x + 3| + |x - 1| - 2 = 0$ is
Solving the Absolute Value Equation
We need to find the number of real solutions for the equation $x|x + 3| + |x - 1| - 2 = 0$. We analyze this equation by dividing the number line into intervals based on the points where the expressions inside the absolute values are zero. These points are $x = -3$ and $x = 1$.
Case 1: $x < -3$
In this interval, $x + 3$ is negative and $x - 1$ is negative. Thus, $|x + 3| = -(x + 3)$ and $|x - 1| = -(x - 1)$.
The equation becomes:
$ x (-(x + 3)) + (-(x - 1)) - 2 = 0 $
$ -x^2 - 3x - x + 1 - 2 = 0 $
$ -x^2 - 4x - 1 = 0 $
$ x^2 + 4x + 1 = 0 $
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$ x = \frac{-4 \pm \sqrt{4^2 - 4(1)(1)}}{2} = \frac{-4 \pm \sqrt{16 - 4}}{2} = \frac{-4 \pm \sqrt{12}}{2} = \frac{-4 \pm 2\sqrt{3}}{2} = -2 \pm \sqrt{3} $
We check if these solutions are in the interval $x < -3$. Since $\sqrt{3} \approx 1.732$:
- $x_1 = -2 + \sqrt{3} \approx -0.268$. This solution is not in the interval $x < -3$.
- $x_2 = -2 - \sqrt{3} \approx -3.732$. This solution is in the interval $x < -3$.
One valid solution found: $x = -2 - \sqrt{3}$.
Case 2: $-3 \le x < 1$
In this interval, $x + 3$ is non-negative and $x - 1$ is negative. Thus, $|x + 3| = x + 3$ and $|x - 1| = -(x - 1)$.
The equation becomes:
$ x(x + 3) + (-(x - 1)) - 2 = 0 $
$ x^2 + 3x - x + 1 - 2 = 0 $
$ x^2 + 2x - 1 = 0 $
Using the quadratic formula:
$ x = \frac{-2 \pm \sqrt{2^2 - 4(1)(-1)}}{2} = \frac{-2 \pm \sqrt{4 + 4}}{2} = \frac{-2 \pm \sqrt{8}}{2} = \frac{-2 \pm 2\sqrt{2}}{2} = -1 \pm \sqrt{2} $
We check if these solutions are in the interval $-3 \le x < 1$. Since $\sqrt{2} \approx 1.414$:
- $x_3 = -1 + \sqrt{2} \approx 0.414$. This solution is in the interval $[-3, 1)$.
- $x_4 = -1 - \sqrt{2} \approx -2.414$. This solution is in the interval $[-3, 1)$.
Two valid solutions found: $x = -1 + \sqrt{2}$ and $x = -1 - \sqrt{2}$.
Case 3: $x \ge 1$
In this interval, $x + 3$ is positive and $x - 1$ is non-negative. Thus, $|x + 3| = x + 3$ and $|x - 1| = x - 1$.
The equation becomes:
$ x(x + 3) + (x - 1) - 2 = 0 $
$ x^2 + 3x + x - 1 - 2 = 0 $
$ x^2 + 4x - 3 = 0 $
Using the quadratic formula:
$ x = \frac{-4 \pm \sqrt{4^2 - 4(1)(-3)}}{2} = \frac{-4 \pm \sqrt{16 + 12}}{2} = \frac{-4 \pm \sqrt{28}}{2} = \frac{-4 \pm 2\sqrt{7}}{2} = -2 \pm \sqrt{7} $
We check if these solutions are in the interval $x \ge 1$. Since $\sqrt{7} \approx 2.646$:
- $x_5 = -2 + \sqrt{7} \approx 0.646$. This solution is not in the interval $x \ge 1$.
- $x_6 = -2 - \sqrt{7} \approx -4.646$. This solution is not in the interval $x \ge 1$.
No valid solutions found in this case.
Total Number of Real Solutions
Summing the valid solutions from the three cases ($x = -2 - \sqrt{3}$, $x = -1 + \sqrt{2}$, $x = -1 - \sqrt{2}$), we find 3 distinct real solutions based on algebraic analysis.
The equation $x|x + 3| + |x - 1| - 2 = 0$ has 4 real solutions.
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