If `""^6C_n +2. ""^6C_(n+1)+""^6 C_(n+2) gt "^8 C_3` then the quadratic equations whose roots are `alpha ,beta` .

Options

A. no common roots
B. 1 common roots
C. 2 common roots
D. imaginary roots

Answer

To solve the inequality \( \binom{6}{n} + 2 \cdot \binom{6}{n+1} + \binom{6}{n+2} > \binom{8}{3} \), we will follow these steps:

### Step 1: Evaluate \( \binom{8}{3} \)
First, we need to calculate \( \binom{8}{3} \):
\[ \binom{8}{3} = \frac{8!}{3!(8-3)!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56 \]

### Step 2: Rewrite the left-hand side using combination identities
Using the identity \( \binom{n}{r} + \binom{n}{r+1} = \binom{n+1}{r+1} \), we can combine terms:
\[ \binom{6}{n} + \binom{6}{n+1} = \binom{7}{n+1} \]
Thus, we can rewrite the left-hand side:
\[ \binom{6}{n} + 2 \cdot \binom{6}{n+1} + \binom{6}{n+2} = \binom{7}{n+1} + \binom{6}{n+2} \]
Now, we apply the identity again:
\[ \binom{6}{n+1} + \binom{6}{n+2} = \binom{7}{n+2} \]
So, we have:
\[ \binom{7}{n+1} + \binom{7}{n+2} = \binom{8}{n+2} \]

### Step 3: Set up the inequality
Now we can set up the inequality:
\[ \binom{8}{n+2} > 56 \]

### Step 4: Solve for \( n+2 \)
Next, we need to find the smallest integer \( n+2 \) such that \( \binom{8}{n+2} > 56 \). We calculate \( \binom{8}{k} \) for \( k = 0, 1, 2, 3, 4, 5, 6, 7, 8 \):
- \( \binom{8}{0} = 1 \)
- \( \binom{8}{1} = 8 \)
- \( \binom{8}{2} = 28 \)
- \( \binom{8}{3} = 56 \)
- \( \binom{8}{4} = 70 \)

From this, we see that \( \binom{8}{4} = 70 > 56 \). Therefore, \( n + 2 \) must be at least 4, which gives us:
\[ n + 2 \geq 4 \implies n \geq 2 \]

### Step 5: Formulate the quadratic equation
The roots of the quadratic equation can be expressed as \( \alpha \) and \( \beta \). The quadratic equation with roots \( \alpha \) and \( \beta \) can be formed using the standard form:
\[ x^2 - (sum \ of \ roots)x + (product \ of \ roots) = 0 \]
Assuming \( \alpha = n \) and \( \beta = n + 1 \), we can express the equation as:
\[ x^2 - (n + (n + 1))x + n(n + 1) = 0 \]
This simplifies to:
\[ x^2 - (2n + 1)x + n(n + 1) = 0 \]

### Final Result
The quadratic equation whose roots are \( \alpha \) and \( \beta \) is:
\[ x^2 - (2n + 1)x + n(n + 1) = 0 \]