Consider the equation of the form `x^(2)+px+q=0`, then number of such equations that have real roots and have coefficients p and q in the set `{1,2,3,4,5,6}` (p may be equal to q) is

Options

A. 18
B. 17
C. 20
D. 19

Answer

To solve the problem, we need to determine the number of quadratic equations of the form \(x^2 + px + q = 0\) that have real roots, where \(p\) and \(q\) are coefficients taken from the set \(\{1, 2, 3, 4, 5, 6\}\).

### Step 1: Understand the condition for real roots
For a quadratic equation \(ax^2 + bx + c = 0\) to have real roots, the discriminant must be non-negative. The discriminant \(\Delta\) for our equation \(x^2 + px + q = 0\) is given by:
\[ \Delta = p^2 - 4q \]
We need \(\Delta \geq 0\) for the roots to be real.

### Step 2: Determine possible values for \(p\) and \(q\)
We will evaluate the values of \(p\) from the set \(\{1, 2, 3, 4, 5, 6\}\) and find corresponding values of \(q\) such that \(p^2 - 4q \geq 0\).

### Step 3: Calculate for each value of \(p\)

1. **For \(p = 1\)**:
\[ 1^2 - 4q \geq 0 \implies 1 - 4q \geq 0 \implies q \leq \frac{1}{4} \]
No valid \(q\) in the set.

2. **For \(p = 2\)**:
\[ 2^2 - 4q \geq 0 \implies 4 - 4q \geq 0 \implies q \leq 1 \]
Valid \(q\): \(1\) (1 solution).

3. **For \(p = 3\)**:
\[ 3^2 - 4q \geq 0 \implies 9 - 4q \geq 0 \implies q \leq \frac{9}{4} = 2.25 \]
Valid \(q\): \(1, 2\) (2 solutions).

4. **For \(p = 4\)**:
\[ 4^2 - 4q \geq 0 \implies 16 - 4q \geq 0 \implies q \leq 4 \]
Valid \(q\): \(1, 2, 3, 4\) (4 solutions).

5. **For \(p = 5\)**:
\[ 5^2 - 4q \geq 0 \implies 25 - 4q \geq 0 \implies q \leq 6.25 \]
Valid \(q\): \(1, 2, 3, 4, 5, 6\) (6 solutions).

6. **For \(p = 6\)**:
\[ 6^2 - 4q \geq 0 \implies 36 - 4q \geq 0 \implies q \leq 9 \]
Valid \(q\): \(1, 2, 3, 4, 5, 6\) (6 solutions).

### Step 4: Total the valid combinations
Now we sum the valid combinations:
- For \(p = 1\): 0 solutions
- For \(p = 2\): 1 solution
- For \(p = 3\): 2 solutions
- For \(p = 4\): 4 solutions
- For \(p = 5\): 6 solutions
- For \(p = 6\): 6 solutions

Total solutions:
\[ 0 + 1 + 2 + 4 + 6 + 6 = 19 \]

### Final Answer
The total number of equations that have real roots is **19**.