Let `alpha, beta ` be ral roots of the quadratic equation`x ^(2)+kx+ (k^(2) +2k-4) =0,` then the minimum value of `alpha ^(z)+beta ^(z)` is equal to :

Options

A. 12
B. `4/9`
C. `16/9`
D. `8/9`

Answer

To find the minimum value of \( \alpha^2 + \beta^2 \) for the quadratic equation \( x^2 + kx + (k^2 + 2k - 4) = 0 \), we will follow these steps:

### Step 1: Determine the Discriminant
The roots \( \alpha \) and \( \beta \) are real if the discriminant \( D \) is greater than or equal to zero. The discriminant for the quadratic equation \( ax^2 + bx + c = 0 \) is given by:
\[ D = b^2 - 4ac \]
For our equation:
- \( a = 1 \)
- \( b = k \)
- \( c = k^2 + 2k - 4 \)

Thus, the discriminant is:
\[ D = k^2 - 4(1)(k^2 + 2k - 4) = k^2 - 4k^2 - 8k + 16 = -3k^2 - 8k + 16 \]
For the roots to be real, we need:
\[ -3k^2 - 8k + 16 \geq 0 \]

### Step 2: Solve the Inequality
Rearranging the inequality:
\[ 3k^2 + 8k - 16 \leq 0 \]
Now we will find the roots of the quadratic equation \( 3k^2 + 8k - 16 = 0 \) using the quadratic formula:
\[ k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-8 \pm \sqrt{8^2 - 4 \cdot 3 \cdot (-16)}}{2 \cdot 3} \]
Calculating the discriminant:
\[ = \frac{-8 \pm \sqrt{64 + 192}}{6} = \frac{-8 \pm \sqrt{256}}{6} = \frac{-8 \pm 16}{6} \]
Thus, the roots are:
\[ k_1 = \frac{8}{6} = \frac{4}{3}, \quad k_2 = \frac{-24}{6} = -4 \]
The quadratic opens upwards (since the coefficient of \( k^2 \) is positive), so the solution to the inequality \( 3k^2 + 8k - 16 \leq 0 \) is:
\[ -4 \leq k \leq \frac{4}{3} \]

### Step 3: Express \( \alpha^2 + \beta^2 \)
Using the identity \( \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \), we need to find \( \alpha + \beta \) and \( \alpha \beta \):
- \( \alpha + \beta = -\frac{b}{a} = -k \)
- \( \alpha \beta = \frac{c}{a} = k^2 + 2k - 4 \)

Thus:
\[ \alpha^2 + \beta^2 = (-k)^2 - 2(k^2 + 2k - 4) = k^2 - 2k^2 - 4k + 8 = -k^2 - 4k + 8 \]

### Step 4: Find the Minimum Value
To find the minimum value of \( -k^2 - 4k + 8 \), we can complete the square:
\[ -k^2 - 4k + 8 = -\left(k^2 + 4k - 8\right) = -\left((k + 2)^2 - 4 - 8\right) = -\left((k + 2)^2 - 12\right) = 12 - (k + 2)^2 \]
The maximum value of \( (k + 2)^2 \) occurs at the endpoints of the interval \( k \in [-4, \frac{4}{3}] \).

### Step 5: Evaluate at the Endpoints
1. For \( k = -4 \):
\[ \alpha^2 + \beta^2 = 12 - (-4 + 2)^2 = 12 - 4 = 8 \]
2. For \( k = \frac{4}{3} \):
\[ \alpha^2 + \beta^2 = 12 - \left(\frac{4}{3} + 2\right)^2 = 12 - \left(\frac{10}{3}\right)^2 = 12 - \frac{100}{9} = \frac{108 - 100}{9} = \frac{8}{9} \]

### Conclusion
The minimum value of \( \alpha^2 + \beta^2 \) occurs at \( k = -4 \) and is:
\[ \boxed{8} \]