The roots of the quadratic equation $x^2 - kx + 12 = 0$ are given as $\alpha$ and $\beta$. Additionally, it is stated that $\alpha - \beta = 1$. We aim to find the value of $k$.

Step 1: Sum and product of the roots

From the standard quadratic equation $x^2 + px + q = 0$, the sum and product of the roots are:

$\alpha + \beta = k \quad \text{(Sum of the roots)}$ $\alpha \beta = 12 \quad \text{(Product of the roots)}$

Step 2: Use the condition $\alpha - \beta = 1$

Let’s use the equations for the sum and difference of roots: $\alpha = \frac{(\alpha + \beta) + (\alpha - \beta)}{2}$ $\beta = \frac{(\alpha + \beta) - (\alpha - \beta)}{2}$

Substitute $\alpha + \beta = k$ and $\alpha - \beta = 1$: $\alpha = \frac{k + 1}{2}, \quad \beta = \frac{k - 1}{2}$

Step 3: Use the product of the roots

The product of the roots $\alpha \beta = 12$ gives: $\left(\frac{k + 1}{2}\right) \left(\frac{k - 1}{2}\right) = 12$

Simplify: $\frac{(k + 1)(k - 1)}{4} = 12$ $\frac{k^2 - 1}{4} = 12$ $k^2 - 1 = 48$ $k^2 = 49$ $k = \pm 7$

Step 4: Verify the solution

If $k = 7$: $\alpha + \beta = 7, \quad \alpha - \beta = 1$ $\alpha = \frac{7 + 1}{2} = 4, \quad \beta = \frac{7 - 1}{2} = 3$ $\alpha \beta = 4 \cdot 3 = 12 \quad \text{(Valid!)}$

If $k = -7$: $\alpha + \beta = -7, \quad \alpha - \beta = 1$ $\alpha = \frac{-7 + 1}{2} = -3, \quad \beta = \frac{-7 - 1}{2} = -4$ $\alpha \beta = -3 \cdot -4 = 12 \quad \text{(Valid!)}$

Final Answer:

The possible values of $k$ are: $\boxed{k = 7 \text{ or } k = -7}$

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Would you like a deeper explanation or any related problems?
Here are 5 related questions:

1. What happens if the roots are complex numbers?
2. How do we derive the quadratic formula for roots?
3. What if $\alpha + \beta$ and $\alpha - \beta$ are both unknown?
4. How do we solve for $k$ if the roots are specified differently?
5. Can this method generalize to polynomials of higher degree?

Tip: Always check the sum and product conditions when solving root-related problems.