If `A={x in R//x^(2)+5|x|+6=0}` then n(A)= . . .

Options

A. 0
B. 4
C. 1
D. 2

Explanation

To solve the equation \( x^2 + 5|x| + 6 = 0 \) and find the number of elements in the set \( A \), we will follow these steps:

### Step 1: Analyze the equation
The equation given is:
\[ x^2 + 5|x| + 6 = 0 \]
We need to consider the absolute value function \( |x| \). This means we will analyze two cases: when \( x \geq 0 \) and when \( x < 0 \).

### Step 2: Case 1: \( x \geq 0 \)
In this case, \( |x| = x \). Thus, the equation becomes:
\[ x^2 + 5x + 6 = 0 \]
Now, we will solve this quadratic equation using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 1, b = 5, c = 6 \).

### Step 3: Calculate the discriminant
First, we calculate the discriminant:
\[ D = b^2 - 4ac = 5^2 - 4 \cdot 1 \cdot 6 = 25 - 24 = 1 \]
Since \( D > 0 \), there are two distinct real roots.

### Step 4: Find the roots
Now, we substitute \( D \) back into the quadratic formula:
\[ x = \frac{-5 \pm \sqrt{1}}{2 \cdot 1} = \frac{-5 \pm 1}{2} \]
Calculating the two roots:
1. \( x_1 = \frac{-5 + 1}{2} = \frac{-4}{2} = -2 \)
2. \( x_2 = \frac{-5 - 1}{2} = \frac{-6}{2} = -3 \)

### Step 5: Analyze the roots
Both roots \( x_1 = -2 \) and \( x_2 = -3 \) are negative. However, since we are in the case where \( x \geq 0 \), these roots are not valid for this case.

### Step 6: Case 2: \( x < 0 \)
In this case, \( |x| = -x \). Thus, the equation becomes:
\[ x^2 - 5x + 6 = 0 \]
We will again use the quadratic formula.

### Step 7: Calculate the discriminant for the second case
\[ D = (-5)^2 - 4 \cdot 1 \cdot 6 = 25 - 24 = 1 \]
Again, since \( D > 0 \), there are two distinct real roots.

### Step 8: Find the roots for the second case
Using the quadratic formula:
\[ x = \frac{5 \pm \sqrt{1}}{2 \cdot 1} = \frac{5 \pm 1}{2} \]
Calculating the two roots:
1. \( x_1 = \frac{5 + 1}{2} = \frac{6}{2} = 3 \)
2. \( x_2 = \frac{5 - 1}{2} = \frac{4}{2} = 2 \)

### Step 9: Analyze the roots for the second case
Both roots \( x_1 = 3 \) and \( x_2 = 2 \) are positive, which means they are not valid for this case where \( x < 0 \).

### Conclusion
Since both cases yield no valid solutions for \( x \), we conclude that there are no values of \( x \) that satisfy the original equation. Therefore, the number of elements in the set \( A \) is:
\[ n(A) = 0 \]

### Final Answer
\[ n(A) = 0 \]