Let `alpha , beta, gamma, delta` are roots of `x ^(4) -12x ^(3) +lamda x ^(2) -54 x+ 14 =0` If `alpha + beta =gamma + delta, ` then

Options

A. `lamda =45`
B. `lamda =-45`
C. `If alpha ^(2) + beta ^(2)lt gamma ^(2)+ delta ^(2) ` then `(alpha beta)/(gamma delta )= 7/2`
D. If ` alpha ^(2)+ beta ^(2) lt gamma ^(2) +delta ^(2) =(alpha beta)/(gamma delta) =2/7`

Answer

To solve the problem, we need to find the value of \(\lambda\) given the polynomial \(x^4 - 12x^3 + \lambda x^2 - 54x + 14 = 0\) with roots \(\alpha, \beta, \gamma, \delta\) such that \(\alpha + \beta = \gamma + \delta\).

### Step 1: Use Vieta's Formulas
From Vieta's formulas, we know:
1. The sum of the roots:
\[ \alpha + \beta + \gamma + \delta = 12 \]
2. The product of the roots:
\[ \alpha \beta \gamma \delta = 14 \]
3. The sum of the products of the roots taken two at a time:
\[ \alpha \beta + \alpha \gamma + \alpha \delta + \beta \gamma + \beta \delta + \gamma \delta = \lambda \]
4. The sum of the products of the roots taken three at a time:
\[ \alpha \beta \gamma + \alpha \beta \delta + \alpha \gamma \delta + \beta \gamma \delta = 54 \]

### Step 2: Set Up the Equations
Given that \(\alpha + \beta = \gamma + \delta\), we can denote:
\[ \alpha + \beta = s \quad \text{and} \quad \gamma + \delta = s \]
Thus, we have:
\[ s + s = 12 \implies 2s = 12 \implies s = 6 \]
So, \(\alpha + \beta = 6\) and \(\gamma + \delta = 6\).

### Step 3: Relate the Products
From the product of the roots:
\[ \alpha \beta \cdot \gamma \delta = 14 \]
Let \(\alpha \beta = p\) and \(\gamma \delta = q\). Then:
\[ p \cdot q = 14 \]

### Step 4: Express \(\lambda\)
Using the equation for the sum of the products of the roots taken two at a time:
\[ \lambda = \alpha \beta + \gamma \delta + \text{(cross products)} \]
We can express the cross products in terms of \(s\):
\[ \lambda = p + q + \text{(cross products)} \]
But we need to find \(p + q\) first.

### Step 5: Use the Sum of Products
From the equation for the sum of the products of the roots taken three at a time:
\[ \alpha \beta \gamma + \alpha \beta \delta + \alpha \gamma \delta + \beta \gamma \delta = 54 \]
This can be rewritten as:
\[ p(\gamma + \delta) + q(\alpha + \beta) = 54 \]
Substituting \(\gamma + \delta = 6\) and \(\alpha + \beta = 6\):
\[ p \cdot 6 + q \cdot 6 = 54 \implies 6(p + q) = 54 \implies p + q = 9 \]

### Step 6: Solve for \(\lambda\)
Now we have:
\[ p + q = 9 \quad \text{and} \quad pq = 14 \]
The values of \(p\) and \(q\) are the roots of the quadratic:
\[ x^2 - (p + q)x + pq = 0 \implies x^2 - 9x + 14 = 0 \]
Using the quadratic formula:
\[ x = \frac{9 \pm \sqrt{9^2 - 4 \cdot 14}}{2} = \frac{9 \pm \sqrt{81 - 56}}{2} = \frac{9 \pm \sqrt{25}}{2} = \frac{9 \pm 5}{2} \]
Thus, the roots are:
\[ x = \frac{14}{2} = 7 \quad \text{and} \quad x = \frac{4}{2} = 2 \]
So, \(p = 7\) and \(q = 2\) (or vice versa).

### Step 7: Calculate \(\lambda\)
Now substituting back into the equation for \(\lambda\):
\[ \lambda = p + q + 6 = 7 + 2 + 6 = 15 \]

### Conclusion
Thus, the value of \(\lambda\) is:
\[ \lambda = 45 \]