If $x$ is a real number, then $x^2 + x + 1$ is

### Concept & Formula The minimum or maximum value of a quadratic expression $ax^2 + bx + c$ can be found by completing the square. $$ax^2 + bx + c = a\left(x + \frac{b}{2a}\right)^2 + \frac{4ac - b^2}{4a}$$ ### Step-by-Step Solution 1. Take the given quadratic expression: $x^2 + x + 1$. 2. To complete the square, add and subtract the square of half the coefficient of $x$, which is $(1/2)^2 = 1/4$. 3. Express it as: $(x^2 + x + 1/4) - 1/4 + 1$. 4. Simplify the expression into a perfect square and a constant: $(x + 1/2)^2 + 3/4$. 5. Since the square of any real number is always non-negative, $(x + 1/2)^2 \ge 0$. 6. Therefore, the entire expression $(x + 1/2)^2 + 3/4 \ge 3/4$. ### Exam Strategy & Shortcut If you know calculus, simply find the derivative and set it to zero to find the extremum. $\frac{d}{dx}(x^2 + x + 1) = 2x + 1 = 0 \Rightarrow x = -1/2$. Substitute $x = -1/2$ back into the expression: $(-1/2)^2 - 1/2 + 1 = 1/4 - 1/2 + 1 = 3/4$. Since the leading coefficient is positive, this is a minimum. ### Common Pitfall Students often try plugging in random positive and negative integer values for $x$ (like 0, 1, -1) and incorrectly deduce the minimum is 1 because they forget to test fractions like $-1/2$. ### Final Answer **Therefore, the correct answer is greater than or equal to 3/4.**