The number of solutions of the equation `x^(2)-4|x|-2=0` is :

Options

A. 1
B. 2
C. 3
D. 4

Answer

To find the number of solutions for the equation \(x^2 - 4|x| - 2 = 0\), we will analyze the equation by considering the two cases for the absolute value function.

### Step 1: Split the equation based on the absolute value
The absolute value function \(|x|\) can be expressed as:
- \(|x| = x\) when \(x \geq 0\)
- \(|x| = -x\) when \(x < 0\)

Thus, we will consider two cases for our equation.

### Step 2: Case 1: \(x \geq 0\)
In this case, we replace \(|x|\) with \(x\):
\[ x^2 - 4x - 2 = 0 \]
Now we will solve this quadratic equation using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \(a = 1\), \(b = -4\), and \(c = -2\).

### Step 3: Calculate the discriminant for Case 1
\[ b^2 - 4ac = (-4)^2 - 4(1)(-2) = 16 + 8 = 24 \]
Since the discriminant is positive, there are two real solutions.

### Step 4: Find the solutions for Case 1
Using the quadratic formula:
\[ x = \frac{4 \pm \sqrt{24}}{2} \]
\[ x = \frac{4 \pm 2\sqrt{6}}{2} \]
\[ x = 2 \pm \sqrt{6} \]
Thus, the solutions are:
\[ x_1 = 2 + \sqrt{6} \quad \text{and} \quad x_2 = 2 - \sqrt{6} \]

### Step 5: Check the validity of solutions for Case 1
- For \(x_1 = 2 + \sqrt{6}\): Since \(\sqrt{6} \approx 2.45\), \(x_1 \approx 4.45\) which is greater than 0. Valid solution.
- For \(x_2 = 2 - \sqrt{6}\): \(x_2 \approx 2 - 2.45 \approx -0.45\) which is less than 0. Not valid in this case.

### Step 6: Case 2: \(x < 0\)
In this case, we replace \(|x|\) with \(-x\):
\[ x^2 + 4x - 2 = 0 \]
Again, we will solve this quadratic equation using the quadratic formula.

### Step 7: Calculate the discriminant for Case 2
\[ b^2 - 4ac = (4)^2 - 4(1)(-2) = 16 + 8 = 24 \]
Again, the discriminant is positive, indicating two real solutions.

### Step 8: Find the solutions for Case 2
Using the quadratic formula:
\[ x = \frac{-4 \pm \sqrt{24}}{2} \]
\[ x = \frac{-4 \pm 2\sqrt{6}}{2} \]
\[ x = -2 \pm \sqrt{6} \]
Thus, the solutions are:
\[ x_3 = -2 + \sqrt{6} \quad \text{and} \quad x_4 = -2 - \sqrt{6} \]

### Step 9: Check the validity of solutions for Case 2
- For \(x_3 = -2 + \sqrt{6}\): Since \(\sqrt{6} \approx 2.45\), \(x_3 \approx -2 + 2.45 \approx 0.45\) which is greater than 0. Not valid in this case.
- For \(x_4 = -2 - \sqrt{6}\): \(x_4 \approx -2 - 2.45 \approx -4.45\) which is less than 0. Valid solution.

### Conclusion
From both cases, we have found:
- From Case 1: One valid solution \(x = 2 + \sqrt{6}\)
- From Case 2: One valid solution \(x = -2 - \sqrt{6}\)

Thus, the total number of valid solutions to the equation \(x^2 - 4|x| - 2 = 0\) is **2**.

---