If all the roots of the equation `x^2 - 3x + k = 0` are real, then k lies in the interval

Options

A. `(-0, -1) `
B. `(-2,2) `
C. `(1, 0) `
D. none of these

Answer

To determine the interval in which \( k \) lies so that all roots of the equation \( x^2 - 3x + k = 0 \) are real, we will follow these steps:

### Step 1: Identify the condition for real roots
For a quadratic equation \( ax^2 + bx + c = 0 \) to have real roots, the discriminant \( D \) must be greater than or equal to zero. The discriminant is given by:
\[ D = b^2 - 4ac \]
In our case, \( a = 1 \), \( b = -3 \), and \( c = k \).

### Step 2: Calculate the discriminant
Substituting the values into the discriminant formula:
\[ D = (-3)^2 - 4(1)(k) = 9 - 4k \]

### Step 3: Set the discriminant greater than or equal to zero
To ensure the roots are real, we set the discriminant \( D \) to be greater than or equal to zero:
\[ 9 - 4k \geq 0 \]

### Step 4: Solve the inequality
Rearranging the inequality gives:
\[ 9 \geq 4k \]
Dividing both sides by 4:
\[ \frac{9}{4} \geq k \]
or equivalently:
\[ k \leq \frac{9}{4} \]

### Step 5: Determine the interval for \( k \)
Since there is no lower bound specified for \( k \), we can say:
\[ k \in (-\infty, \frac{9}{4}] \]

### Conclusion
Thus, the interval in which \( k \) lies for all roots of the equation \( x^2 - 3x + k = 0 \) to be real is:
\[ (-\infty, \frac{9}{4}] \]