If `alpha, beta` are the roots of the equation `x^(2)+7x+12=0`, then the equation whose roots are `(alpha+beta)^(2)` and `(alpha-beta)^(2)` is

Options

A. `x^(2)+50x+49=0`
B. `x^(2)-50x+49=0`
C. `x^(2)-50x-49=0`
D. `x^(2)+12x+7=0`

Answer

To solve the problem, we will follow these steps:

### Step 1: Find the roots of the given quadratic equation
The given quadratic equation is:
\[ x^2 + 7x + 12 = 0 \]

To find the roots, we can factor the equation:
We need two numbers that multiply to \(12\) (the constant term) and add up to \(7\) (the coefficient of \(x\)). The numbers \(3\) and \(4\) satisfy this condition.

Thus, we can factor the equation as:
\[ (x + 3)(x + 4) = 0 \]

Setting each factor to zero gives us the roots:
\[ x + 3 = 0 \quad \Rightarrow \quad x = -3 \]
\[ x + 4 = 0 \quad \Rightarrow \quad x = -4 \]

So, the roots are:
\[ \alpha = -4, \quad \beta = -3 \]

### Step 2: Calculate \( \alpha + \beta \) and \( \alpha - \beta \)
Now, we calculate \( \alpha + \beta \) and \( \alpha - \beta \):
\[ \alpha + \beta = -4 + (-3) = -7 \]
\[ \alpha - \beta = -4 - (-3) = -4 + 3 = -1 \]

### Step 3: Calculate \( (\alpha + \beta)^2 \) and \( (\alpha - \beta)^2 \)
Next, we find the squares of these sums:
\[ (\alpha + \beta)^2 = (-7)^2 = 49 \]
\[ (\alpha - \beta)^2 = (-1)^2 = 1 \]

### Step 4: Form the new quadratic equation
Let the roots of the new quadratic equation be \( k = (\alpha + \beta)^2 = 49 \) and \( k' = (\alpha - \beta)^2 = 1 \).

The general form of a quadratic equation with roots \( k \) and \( k' \) is:
\[ x^2 - (k + k')x + k \cdot k' = 0 \]

Calculating \( k + k' \) and \( k \cdot k' \):
\[ k + k' = 49 + 1 = 50 \]
\[ k \cdot k' = 49 \cdot 1 = 49 \]

Thus, the quadratic equation is:
\[ x^2 - 50x + 49 = 0 \]

### Conclusion
Therefore, the quadratic equation whose roots are \( (\alpha + \beta)^2 \) and \( (\alpha - \beta)^2 \) is:
\[ \boxed{x^2 - 50x + 49 = 0} \]

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