`x^2+2x+5gt0`

Options

A. (A) `(-oo,oo)`
B. (B) `[-oo,oo]`
C. (C) `(-oo,1)uu(3oo)`
D. (D) `(-oo,-2)uu(-5,oo)`

Answer

To solve the inequality \(x^2 + 2x + 5 > 0\), we can follow these steps:

### Step 1: Rewrite the quadratic expression
We start with the expression:
\[ x^2 + 2x + 5 \]
We can complete the square for the quadratic part.

### Step 2: Completing the square
To complete the square, we take the coefficient of \(x\) (which is 2), halve it (getting 1), and square it (getting 1). Thus, we can rewrite the expression:
\[ x^2 + 2x + 1 - 1 + 5 \]
This simplifies to:
\[ (x + 1)^2 + 4 \]

### Step 3: Analyze the completed square
Now, we have:
\[ (x + 1)^2 + 4 > 0 \]
Since \((x + 1)^2\) is always non-negative (it is a square), the smallest value it can take is 0. Thus:
\[ (x + 1)^2 \geq 0 \]
Adding 4 to this gives:
\[ (x + 1)^2 + 4 \geq 4 \]
This means:
\[ (x + 1)^2 + 4 > 0 \]
is always true for all real values of \(x\).

### Step 4: Conclusion
Since the expression \(x^2 + 2x + 5\) is always greater than 0 for all real numbers \(x\), the solution to the inequality is:
\[ \text{All real numbers } x \in (-\infty, \infty) \]