Under which one of the following conditions will the quadatic equation `x^(2) + mx + 2 =0` always have real roots ?

Options

A. `2sqrt(3) le m^(2) lt 8`
B. `sqrt(3) le m^(2) lt 4`
C. `m^(2) ge 8`
D. `m^(2) le sqrt(3)`

Answer

To determine the conditions under which the quadratic equation \( x^2 + mx + 2 = 0 \) always has real roots, we need to analyze the discriminant of the quadratic equation.

### Step-by-Step Solution:

1. **Identify the coefficients**:
The given quadratic equation is in the standard form \( Ax^2 + Bx + C = 0 \).
Here, \( A = 1 \), \( B = m \), and \( C = 2 \).

2. **Write the formula for the discriminant**:
The discriminant \( D \) of a quadratic equation is given by:
\[ D = B^2 - 4AC \]
For the equation to have real roots, the discriminant must be greater than or equal to zero:
\[ D \geq 0 \]

3. **Substitute the coefficients into the discriminant formula**:
Substitute \( A \), \( B \), and \( C \) into the discriminant:
\[ D = m^2 - 4 \cdot 1 \cdot 2 \]
Simplifying this gives:
\[ D = m^2 - 8 \]

4. **Set the discriminant greater than or equal to zero**:
To ensure real roots, we set up the inequality:
\[ m^2 - 8 \geq 0 \]

5. **Solve the inequality**:
Rearranging the inequality gives:
\[ m^2 \geq 8 \]
Taking the square root of both sides, we find:
\[ |m| \geq \sqrt{8} \quad \text{or} \quad |m| \geq 2\sqrt{2} \]
This means:
\[ m \leq -2\sqrt{2} \quad \text{or} \quad m \geq 2\sqrt{2} \]

6. **Identify the correct option**:
Among the options provided, we need to find which one corresponds to \( m^2 \geq 8 \). The option that states \( m^2 \) should be greater than or equal to 8 is option C.

### Conclusion:
Thus, the condition under which the quadratic equation \( x^2 + mx + 2 = 0 \) always has real roots is:
\[ \text{Option C: } m^2 \geq 8 \]