Question

Column A: \(x^2 + 2x + 1\)
Column B: \(x^2\)

• A. The quantity in Column A is greater.
• B. The quantity in Column B is greater.
• C. The two quantities are equal.
• D. The relationship cannot be determined from the information given.

Hint:

When comparing algebraic expressions, testing a few strategic numbers is a powerful technique. Always try a positive number (like 1 or 2), a negative number (like -1 or -2), zero (0), and a fraction between 0 and 1 if applicable. If you get different results, the answer is (D).

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
This problem asks us to compare two algebraic expressions. Since the value of \(x\) is not specified, we should test different values for \(x\) (positive, negative, and zero) to see if the relationship between the columns is constant.
Step 2: Key Formula or Approach:
First, simplify the expression in Column A. The expression \(x^2 + 2x + 1\) is a perfect square trinomial, which can be factored as \((x+1)^2\). So we are comparing \((x+1)^2\) with \(x^2\). A simpler way is to subtract \(x^2\) from both columns and compare \(2x+1\) with 0.
Step 3: Detailed Explanation:
Let's compare \(x^2 + 2x + 1\) and \(x^2\). We can simplify the comparison by subtracting \(x^2\) from both quantities. The comparison then becomes between \(2x + 1\) and 0.
The relationship depends on the value of \(x\):
\begin{itemize} \item Case 1: Test a positive value for \(x\).
Let \(x = 1\). Column A: \(1^2 + 2(1) + 1 = 1 + 2 + 1 = 4\). Column B: \(1^2 = 1\). In this case, Column A>Column B. (Because \(2(1)+1 = 3>0\)). \item Case 2: Test a negative value for \(x\).
Let \(x = -2\). Column A: \((-2)^2 + 2(-2) + 1 = 4 - 4 + 1 = 1\). Column B: \((-2)^2 = 4\). In this case, Column A<Column B. (Because \(2(-2)+1 = -3<0\)). \item Case 3: Test a special value.
Let \(x = -1/2\). Column A: \((-1/2)^2 + 2(-1/2) + 1 = 1/4 - 1 + 1 = 1/4\). Column B: \((-1/2)^2 = 1/4\). In this case, Column A = Column B. (Because \(2(-1/2)+1 = 0\)). \end{itemize} Step 4: Final Answer:
Because we found cases where A>B, A<B, and A = B, the relationship is not fixed. Therefore, the relationship cannot be determined from the information given.