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

Options

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

Answer

To determine the number of equations of the form \(x^2 + ax + b = 0\) that have real roots, where \(a\) and \(b\) are chosen from the set \(\{1, 2, 3, 4, 5, 6\}\), we need to use the condition for real roots, which is based on the discriminant of the quadratic equation.

### Step-by-Step Solution:

1. **Identify the Discriminant**:
The discriminant \(D\) of the quadratic equation \(x^2 + ax + b = 0\) is given by:
\[ D = a^2 - 4b \]
For the roots to be real, the discriminant must be non-negative:
\[ D \geq 0 \quad \Rightarrow \quad a^2 - 4b \geq 0 \quad \Rightarrow \quad a^2 \geq 4b \]

2. **Determine Possible Values of \(a\) and \(b\)**:
Since \(a\) and \(b\) can take values from the set \(\{1, 2, 3, 4, 5, 6\}\), we will evaluate each possible value of \(a\) and find the corresponding values of \(b\) that satisfy the inequality \(a^2 \geq 4b\).

3. **Calculate for Each Value of \(a\)**:
- **If \(a = 1\)**:
\[ 1^2 \geq 4b \quad \Rightarrow \quad 1 \geq 4b \quad \Rightarrow \quad b \leq \frac{1}{4} \quad \Rightarrow \quad \text{No valid } b \]
- **If \(a = 2\)**:
\[ 2^2 \geq 4b \quad \Rightarrow \quad 4 \geq 4b \quad \Rightarrow \quad b \leq 1 \quad \Rightarrow \quad b = 1 \quad (1 \text{ valid pair: } (2, 1)) \]
- **If \(a = 3\)**:
\[ 3^2 \geq 4b \quad \Rightarrow \quad 9 \geq 4b \quad \Rightarrow \quad b \leq \frac{9}{4} \quad \Rightarrow \quad b = 1, 2 \quad (2 \text{ valid pairs: } (3, 1), (3, 2)) \]
- **If \(a = 4\)**:
\[ 4^2 \geq 4b \quad \Rightarrow \quad 16 \geq 4b \quad \Rightarrow \quad b \leq 4 \quad \Rightarrow \quad b = 1, 2, 3, 4 \quad (4 \text{ valid pairs: } (4, 1), (4, 2), (4, 3), (4, 4)) \]
- **If \(a = 5\)**:
\[ 5^2 \geq 4b \quad \Rightarrow \quad 25 \geq 4b \quad \Rightarrow \quad b \leq 6.25 \quad \Rightarrow \quad b = 1, 2, 3, 4, 5, 6 \quad (6 \text{ valid pairs: } (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)) \]
- **If \(a = 6\)**:
\[ 6^2 \geq 4b \quad \Rightarrow \quad 36 \geq 4b \quad \Rightarrow \quad b \leq 9 \quad \Rightarrow \quad b = 1, 2, 3, 4, 5, 6 \quad (6 \text{ valid pairs: } (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)) \]

4. **Count the Valid Pairs**:
Now we sum up the valid pairs:
- For \(a = 1\): 0 pairs
- For \(a = 2\): 1 pair
- For \(a = 3\): 2 pairs
- For \(a = 4\): 4 pairs
- For \(a = 5\): 6 pairs
- For \(a = 6\): 6 pairs

Total valid pairs = \(0 + 1 + 2 + 4 + 6 + 6 = 19\).

### Final Answer:
Thus, the total number of equations that have real roots is **19**.