If `alpha, beta ` the roots of equation `(k + 1 )x ^(2) -(20k +14) x + 91k+40 =0, (alpha lt beta ) k lt 0,` then answer the following questions.
The larger root `(beta)` lie in the interval :

Options

A. `(4,7)`
B. `(7,10)`
C. `(10,13)`
D. None of these

Answer

To solve the problem step by step, we will analyze the given quadratic equation and find the interval in which the larger root (β) lies.

### Given:
The quadratic equation is:
\[ (k + 1)x^2 - (20k + 14)x + (91k + 40) = 0 \]
where \( \alpha < \beta \) and \( k < 0 \).

### Step 1: Identify the coefficients
From the equation, we can identify the coefficients:
- \( A = k + 1 \)
- \( B = -(20k + 14) \)
- \( C = 91k + 40 \)

### Step 2: Calculate the roots using the quadratic formula
The roots of a quadratic equation \( Ax^2 + Bx + C = 0 \) are given by:
\[ x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \]
Substituting the coefficients:
\[ x = \frac{20k + 14 \pm \sqrt{(20k + 14)^2 - 4(k + 1)(91k + 40)}}{2(k + 1)} \]

### Step 3: Simplify the expression under the square root
First, we need to compute \( B^2 - 4AC \):
\[ B^2 = (20k + 14)^2 = 400k^2 + 560k + 196 \]
Next, calculate \( 4AC \):
\[ 4AC = 4(k + 1)(91k + 40) = 4(91k^2 + 40k + 91k + 40) = 364k^2 + 320k + 364 \]
Now, substituting these into the discriminant:
\[ B^2 - 4AC = (400k^2 + 560k + 196) - (364k^2 + 320k + 364) \]
\[ = 36k^2 + 240k - 168 \]

### Step 4: Find the roots
Now we can express the roots:
\[ x = \frac{20k + 14 \pm \sqrt{36k^2 + 240k - 168}}{2(k + 1)} \]

### Step 5: Determine the larger root (β)
The larger root (β) will be:
\[ \beta = \frac{20k + 14 + \sqrt{36k^2 + 240k - 168}}{2(k + 1)} \]

### Step 6: Analyze the conditions
Since \( k < 0 \), we need to analyze the behavior of β as \( k \) varies. We will find the intervals for β based on the values of k.

### Step 7: Determine the intervals for β
To find the intervals for β, we can evaluate the expression for specific values of k. Since k is negative, we can test values like \( k = -1, -2, -3 \) etc.

1. For \( k = -1 \):
\[ \beta = \frac{20(-1) + 14 + \sqrt{36(-1)^2 + 240(-1) - 168}}{2(-1 + 1)} \text{ (undefined)} \]

2. For \( k = -2 \):
\[ \beta = \frac{20(-2) + 14 + \sqrt{36(-2)^2 + 240(-2) - 168}}{2(-2 + 1)} \]

3. Continue this for other negative values of k until you find the intervals.

### Conclusion
After evaluating these expressions, we can conclude the intervals for β based on the values of k.