If ` a lt 0,` then the value of x satisfying `x ^(2)-2a|x-a| -3a ^(2)=0` is/are

Options

A. `a (1- sqrt2)`
B. `a (1+ sqrt2)`
C. `a (-1 -sqrt6 )`
D. `a(-1+sqrt6)`

Answer

To solve the equation \( x^2 - 2a|x-a| - 3a^2 = 0 \) given that \( a < 0 \), we can break it down into two cases based on the definition of the absolute value.

### Step 1: Identify Cases for Absolute Value
The absolute value \( |x-a| \) can be expressed in two cases:
- **Case 1:** \( x - a \geq 0 \) (i.e., \( x \geq a \)) implies \( |x-a| = x-a \)
- **Case 2:** \( x - a < 0 \) (i.e., \( x < a \)) implies \( |x-a| = -(x-a) = a-x \)

### Step 2: Solve Case 1
In Case 1, substituting \( |x-a| = x-a \) into the equation gives:
\[ x^2 - 2a(x-a) - 3a^2 = 0 \]
Simplifying this:
\[ x^2 - 2ax + 2a^2 - 3a^2 = 0 \]
\[ x^2 - 2ax - a^2 = 0 \]

### Step 3: Apply the Quadratic Formula
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1, b = -2a, c = -a^2 \):
\[ x = \frac{-(-2a) \pm \sqrt{(-2a)^2 - 4 \cdot 1 \cdot (-a^2)}}{2 \cdot 1} \]
\[ x = \frac{2a \pm \sqrt{4a^2 + 4a^2}}{2} \]
\[ x = \frac{2a \pm \sqrt{8a^2}}{2} \]
\[ x = \frac{2a \pm 2a\sqrt{2}}{2} \]
\[ x = a(1 \pm \sqrt{2}) \]

### Step 4: Analyze Roots from Case 1
The roots from Case 1 are:
1. \( x_1 = a(1 + \sqrt{2}) \)
2. \( x_2 = a(1 - \sqrt{2}) \)

Since \( a < 0 \):
- \( x_1 = a(1 + \sqrt{2}) < 0 \) (as \( 1 + \sqrt{2} > 0 \))
- \( x_2 = a(1 - \sqrt{2}) \) (Here, \( 1 - \sqrt{2} < 0 \) since \( \sqrt{2} > 1 \))

Thus, both roots are negative, but we need to check if they satisfy \( x \geq a \).

### Step 5: Solve Case 2
In Case 2, substituting \( |x-a| = a-x \) gives:
\[ x^2 - 2a(a-x) - 3a^2 = 0 \]
Simplifying this:
\[ x^2 + 2a x - 2a^2 - 3a^2 = 0 \]
\[ x^2 + 2a x - 5a^2 = 0 \]

### Step 6: Apply the Quadratic Formula for Case 2
Using the quadratic formula:
Here, \( a = 1, b = 2a, c = -5a^2 \):
\[ x = \frac{-2a \pm \sqrt{(2a)^2 - 4 \cdot 1 \cdot (-5a^2)}}{2 \cdot 1} \]
\[ x = \frac{-2a \pm \sqrt{4a^2 + 20a^2}}{2} \]
\[ x = \frac{-2a \pm \sqrt{24a^2}}{2} \]
\[ x = \frac{-2a \pm 2a\sqrt{6}}{2} \]
\[ x = -a(1 \mp \sqrt{6}) \]

### Step 7: Analyze Roots from Case 2
The roots from Case 2 are:
1. \( x_3 = -a(1 + \sqrt{6}) \)
2. \( x_4 = -a(1 - \sqrt{6}) \)

Since \( a < 0 \):
- \( x_3 = -a(1 + \sqrt{6}) > 0 \) (as \( 1 + \sqrt{6} > 0 \))
- \( x_4 = -a(1 - \sqrt{6}) < 0 \) (as \( 1 - \sqrt{6} < 0 \))

### Conclusion
The valid solutions satisfying \( x \geq a \) are:
1. From Case 1: \( x = a(1 - \sqrt{2}) \) (valid since \( x < a \))
2. From Case 2: \( x = -a(1 + \sqrt{6}) \) (valid since \( x > 0 \))

Thus, the values of \( x \) satisfying the equation are:
- \( x = a(1 - \sqrt{2}) \)
- \( x = -a(1 + \sqrt{6}) \)