If `.^(6)C_(k) + 2* .^(6)C_(k+1) + .^(6)C_(k+2) gt .^(8)C_(3)` then the quadratic equation whose roots are `alpha, beta and alpha^(k-1), beta^(k-1)` have

Options

A. no common root
B. one common root
C. both common roots
D. imaginary roots

Explanation

To solve the given problem step by step, we will analyze the expression and find the nature of the roots of the quadratic equation.

### Step 1: Analyze the Given Inequality
We start with the inequality:
\[ \binom{6}{k} + 2 \cdot \binom{6}{k+1} + \binom{6}{k+2} > \binom{8}{3} \]

### Step 2: Simplify the Left Side
Using the property of combinations:
\[ \binom{n}{r} + \binom{n}{r+1} = \binom{n+1}{r+1} \]
we can group the terms:
\[ \binom{6}{k} + \binom{6}{k+1} = \binom{7}{k+1} \]
Thus, we can rewrite the left side:
\[ \binom{7}{k+1} + \binom{6}{k+2} \]
Now, applying the property again:
\[ \binom{6}{k+1} + \binom{6}{k+2} = \binom{7}{k+2} \]
So, we have:
\[ \binom{7}{k+1} + \binom{7}{k+2} = \binom{8}{k+2} \]
Thus, the inequality becomes:
\[ \binom{8}{k+2} > \binom{8}{3} \]

### Step 3: Analyze the Combinations
We know that:
\[ \binom{8}{4} > \binom{8}{3} \]
This means:
\[ k+2 < 4 \]
So, we can solve for \(k\):
\[ k < 2 \]

### Step 4: Determine the Value of k
Since \(k\) must be a non-negative integer, the possible values for \(k\) are:
- \(k = 0\) or \(k = 1\)

### Step 5: Roots of the Quadratic Equation
The roots of the quadratic equation are given as:
\[ \alpha, \beta, \alpha^{k-1}, \beta^{k-1} \]
For \(k = 2\):
- The roots become \(\alpha, \beta, \alpha^{1}, \beta^{1}\) which simplifies to \(\alpha, \beta, \alpha, \beta\).

### Step 6: Nature of the Roots
Since both pairs of roots are the same, the quadratic equation has repeated roots.

### Conclusion
Thus, the nature of the roots of the quadratic equation is that they are both common roots.