If the equation `|x^2-5x + 6|-lambda x+7 lambda=0` has exactly 3 distinct solutions then `lambda` is equal to

Options

A. `-7+sqrt(23)`
B. `-9+4sqrt(5)`
C. `-7-sqrt(23)`
D. `-9-4sqrt(5)`

Explanation

To solve the equation \( |x^2 - 5x + 6| - \lambda x + 7\lambda = 0 \) for the value of \( \lambda \) such that the equation has exactly 3 distinct solutions, we can follow these steps:

### Step 1: Analyze the expression inside the absolute value
The expression inside the absolute value is \( x^2 - 5x + 6 \). We can factor this quadratic expression:
\[ x^2 - 5x + 6 = (x - 2)(x - 3) \]
This means the roots of the quadratic are \( x = 2 \) and \( x = 3 \).

### Step 2: Determine the intervals for the absolute value
The expression \( |x^2 - 5x + 6| \) will change at the points where \( x^2 - 5x + 6 = 0 \), which are \( x = 2 \) and \( x = 3 \). Therefore, we can consider the following intervals:
1. \( x < 2 \)
2. \( 2 \leq x < 3 \)
3. \( x \geq 3 \)

### Step 3: Write the piecewise function
For each interval, we can express the absolute value:
- For \( x < 2 \): \( |x^2 - 5x + 6| = -(x^2 - 5x + 6) = -x^2 + 5x - 6 \)
- For \( 2 \leq x < 3 \): \( |x^2 - 5x + 6| = x^2 - 5x + 6 \)
- For \( x \geq 3 \): \( |x^2 - 5x + 6| = x^2 - 5x + 6 \)

### Step 4: Set up the equations for each interval
We need to solve the equation \( |x^2 - 5x + 6| - \lambda x + 7\lambda = 0 \) in each interval.

1. **For \( x < 2 \)**:
\[ -x^2 + 5x - 6 - \lambda x + 7\lambda = 0 \implies -x^2 + (5 - \lambda)x + (7\lambda - 6) = 0 \]

2. **For \( 2 \leq x < 3 \)**:
\[ x^2 - 5x + 6 - \lambda x + 7\lambda = 0 \implies x^2 + (-5 - \lambda)x + (6 + 7\lambda) = 0 \]

3. **For \( x \geq 3 \)**:
\[ x^2 - 5x + 6 - \lambda x + 7\lambda = 0 \implies x^2 + (-5 - \lambda)x + (6 + 7\lambda) = 0 \]

### Step 5: Conditions for exactly 3 distinct solutions
For the overall equation to have exactly 3 distinct solutions, one of the quadratic equations must have a double root, and the other two must have distinct roots. This means we need to set the discriminant of one of the quadratic equations to zero.

### Step 6: Set the discriminant to zero
Let's consider the equation for \( x < 2 \):
\[ D = (5 - \lambda)^2 - 4(-1)(7\lambda - 6) = 0 \]
Expanding this gives:
\[ (5 - \lambda)^2 + 4(7\lambda - 6) = 0 \]
\[ 25 - 10\lambda + \lambda^2 + 28\lambda - 24 = 0 \]
\[ \lambda^2 + 18\lambda + 1 = 0 \]

### Step 7: Solve for \( \lambda \)
Using the quadratic formula:
\[ \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-18 \pm \sqrt{18^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} \]
\[ = \frac{-18 \pm \sqrt{324 - 4}}{2} = \frac{-18 \pm \sqrt{320}}{2} = \frac{-18 \pm 8\sqrt{5}}{2} \]
\[ = -9 \pm 4\sqrt{5} \]

### Conclusion
Thus, the values of \( \lambda \) that satisfy the condition of having exactly 3 distinct solutions are:
\[ \lambda = -9 + 4\sqrt{5} \quad \text{or} \quad \lambda = -9 - 4\sqrt{5} \]