If the roots of the equation `4B^(2)+ lambda B-2 =0` are of the from `k/(k+1)and (k+1)/(k+2),` then what is the value of `lambda`?

Options

A. 2k
B. 7
C. 2
D. `k+1`

Explanation

To find the value of \( \lambda \) in the equation \( 4B^2 + \lambda B - 2 = 0 \) given that the roots are of the form \( \frac{k}{k+1} \) and \( \frac{k+1}{k+2} \), we can follow these steps:

### Step 1: Identify the roots
The roots of the equation are given as:
\[ r_1 = \frac{k}{k+1}, \quad r_2 = \frac{k+1}{k+2} \]

### Step 2: Use Vieta's formulas
According to Vieta's formulas, for a quadratic equation \( ax^2 + bx + c = 0 \):
- The sum of the roots \( r_1 + r_2 = -\frac{b}{a} \)
- The product of the roots \( r_1 \cdot r_2 = \frac{c}{a} \)

Here, \( a = 4 \), \( b = \lambda \), and \( c = -2 \).

### Step 3: Calculate the sum of the roots
Calculating the sum of the roots:
\[ r_1 + r_2 = \frac{k}{k+1} + \frac{k+1}{k+2} \]
To add these fractions, we find a common denominator:
\[ = \frac{k(k+2) + (k+1)(k+1)}{(k+1)(k+2)} = \frac{k^2 + 2k + k^2 + 2k + 1}{(k+1)(k+2)} = \frac{2k^2 + 4k + 1}{(k+1)(k+2)} \]

Setting this equal to \( -\frac{\lambda}{4} \):
\[ \frac{2k^2 + 4k + 1}{(k+1)(k+2)} = -\frac{\lambda}{4} \]

### Step 4: Calculate the product of the roots
Now calculating the product of the roots:
\[ r_1 \cdot r_2 = \frac{k}{k+1} \cdot \frac{k+1}{k+2} = \frac{k}{k+2} \]
Setting this equal to \( \frac{-2}{4} = -\frac{1}{2} \):
\[ \frac{k}{k+2} = -\frac{1}{2} \]

### Step 5: Solve for \( k \)
Cross-multiplying gives:
\[ 2k = -k - 2 \implies 3k = -2 \implies k = -\frac{2}{3} \]

### Step 6: Substitute \( k \) back to find \( \lambda \)
Now substitute \( k = -\frac{2}{3} \) back into the equation for the sum of the roots:
\[ \frac{2(-\frac{2}{3})^2 + 4(-\frac{2}{3}) + 1}{(-\frac{2}{3}+1)(-\frac{2}{3}+2)} = -\frac{\lambda}{4} \]
Calculating the left-hand side:
\[ = \frac{2 \cdot \frac{4}{9} - \frac{8}{3} + 1}{\frac{1}{3} \cdot \frac{4}{3}} = \frac{\frac{8}{9} - \frac{24}{9} + \frac{9}{9}}{\frac{4}{9}} = \frac{\frac{-7}{9}}{\frac{4}{9}} = -\frac{7}{4} \]
Setting this equal to \( -\frac{\lambda}{4} \):
\[ -\frac{7}{4} = -\frac{\lambda}{4} \implies \lambda = 7 \]

### Final Answer
Thus, the value of \( \lambda \) is:
\[ \boxed{7} \]