The set of all solutions of the inequation `x^(2) - 2x +5 le 0` in R is

Options

A. a) `R-(-infty, -5)`
B. b) `R- (5, infty)`
C. c) `phi`
D. d) `R-(-infty, -4)`

Answer

To solve the inequation \( x^2 - 2x + 5 \leq 0 \), we will follow these steps:

### Step 1: Identify the quadratic expression
The given inequation is:
\[ x^2 - 2x + 5 \leq 0 \]
This is a quadratic expression in the standard form \( ax^2 + bx + c \) where:
- \( a = 1 \)
- \( b = -2 \)
- \( c = 5 \)

### Step 2: Calculate the discriminant
The discriminant \( D \) of a quadratic equation \( ax^2 + bx + c \) is given by:
\[ D = b^2 - 4ac \]
Substituting the values of \( a \), \( b \), and \( c \):
\[ D = (-2)^2 - 4 \cdot 1 \cdot 5 = 4 - 20 = -16 \]

### Step 3: Analyze the discriminant
Since the discriminant \( D = -16 \) is less than 0, this indicates that the quadratic equation has no real roots.

### Step 4: Determine the nature of the quadratic expression
A quadratic expression \( ax^2 + bx + c \) opens upwards (as \( a > 0 \)). Since it has no real roots and opens upwards, the expression \( x^2 - 2x + 5 \) is always positive for all real values of \( x \).

### Step 5: Conclusion about the inequation
Since \( x^2 - 2x + 5 \) is always greater than 0, it can never be less than or equal to 0. Therefore, the solution set for the inequation \( x^2 - 2x + 5 \leq 0 \) is the null set.

### Final Answer
The set of all solutions of the inequation \( x^2 - 2x + 5 \leq 0 \) in \( \mathbb{R} \) is:
\[ \emptyset \quad \text{(null set)} \]

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