Suppose that x = -2 is the only number that satisfies 2^(x)^2 + mathrm{Bx + 11/2} = sqrt(8). What is...

1. Translate the problem requirements: We need to find the value of B such that \(\mathrm{x = -2}\) is the ONLY solution to the exponential equation. This means when we substitute \(\mathrm{x = -2}\), the equation must be satisfied, AND the quadratic expression in the exponent must have exactly one solution.
2. Convert both sides to the same base: Transform the equation so both sides have the same base, allowing us to equate the exponents directly.
3. Apply the constraint that x = -2 satisfies the equation: Substitute \(\mathrm{x = -2}\) into the simplified equation to establish a relationship involving B.
4. Use the uniqueness condition: Since \(\mathrm{x = -2}\) is the ONLY solution, the quadratic equation we derived must have exactly one solution, meaning it's a perfect square with \(\mathrm{x = -2}\) as its repeated root.

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what this problem is really asking. We have an exponential equation where \(\mathrm{x = -2}\) is the ONLY number that works. This tells us two important things:

First, when we plug in \(\mathrm{x = -2}\), the equation must be true. Second, and this is the key insight, \(\mathrm{x = -2}\) must be the only value that makes the equation work - no other number should satisfy it.

Think of it like finding the only key that opens a specific lock. Not only must our key (\(\mathrm{x = -2}\)) open the lock, but it must be the ONLY key that works.

Process Skill: TRANSLATE - Converting the phrase "only number that satisfies" into the mathematical constraint that we need exactly one solution

2. Convert both sides to the same base

To solve exponential equations, we want both sides to have the same base so we can compare the exponents directly. It's like comparing two towers - if they're built with the same sized blocks, we can just count the blocks to see which is taller.

Let's look at our equation: \(\mathrm{2^{x^2 + Bx + \frac{11}{2}} = \sqrt{8}}\)

The left side already has base 2. For the right side, let's rewrite \(\sqrt{8}\) using base 2: \(\sqrt{8} = \sqrt{2^3} = (2^3)^{\frac{1}{2}} = 2^{\frac{3}{2}}\)

So our equation becomes: \(\mathrm{2^{x^2 + Bx + \frac{11}{2}} = 2^{\frac{3}{2}}}\)

Now that both sides have the same base, we can set the exponents equal: \(\mathrm{x^2 + Bx + \frac{11}{2} = \frac{3}{2}}\)

3. Apply the constraint that x = -2 satisfies the equation

Since we know \(\mathrm{x = -2}\) must work in our equation, let's substitute it into our simplified equation:

\(\mathrm{(-2)^2 + B(-2) + \frac{11}{2} = \frac{3}{2}}\)

Let's calculate step by step: \(\mathrm{4 - 2B + \frac{11}{2} = \frac{3}{2}}\)

To make this easier to work with, let's convert 4 to halves: \(\mathrm{4 = \frac{8}{2}}\)

\(\mathrm{\frac{8}{2} - 2B + \frac{11}{2} = \frac{3}{2}}\)

Combining the fractions on the left: \(\mathrm{\frac{8 + 11}{2} - 2B = \frac{3}{2}}\) \(\mathrm{\frac{19}{2} - 2B = \frac{3}{2}}\)

Subtracting \(\frac{19}{2}\) from both sides: \(\mathrm{-2B = \frac{3}{2} - \frac{19}{2} = \frac{-16}{2} = -8}\)

Therefore: \(\mathrm{B = 4}\)

4. Use the uniqueness condition

Wait - we need to verify that \(\mathrm{x = -2}\) is indeed the ONLY solution. Let's check our quadratic equation: \(\mathrm{x^2 + Bx + \frac{11}{2} = \frac{3}{2}}\)

Rearranging: \(\mathrm{x^2 + Bx + \frac{11}{2} - \frac{3}{2} = 0}\) \(\mathrm{x^2 + Bx + \frac{8}{2} = 0}\) \(\mathrm{x^2 + Bx + 4 = 0}\)

With \(\mathrm{B = 4}\), this becomes: \(\mathrm{x^2 + 4x + 4 = 0}\)

This factors as: \(\mathrm{(x + 2)^2 = 0}\)

Perfect! This equation has exactly one solution: \(\mathrm{x = -2}\) (with multiplicity 2). This confirms that \(\mathrm{x = -2}\) is indeed the ONLY value that satisfies our original equation.

Process Skill: APPLY CONSTRAINTS - Using the uniqueness condition to verify our answer is correct

Final Answer

The value of B is 4.

Verification: When \(\mathrm{B = 4}\), our quadratic becomes \(\mathrm{(x + 2)^2 = 0}\), which has exactly one solution \(\mathrm{x = -2}\), satisfying both conditions in the problem.

The answer is C.

Common Faltering Points

Errors while devising the approach

1. Missing the uniqueness constraint

Many students focus only on the fact that \(\mathrm{x = -2}\) satisfies the equation but completely overlook the critical word "ONLY." They substitute \(\mathrm{x = -2}\), solve for B, and think they're done. However, the constraint that \(\mathrm{x = -2}\) is the ONLY solution means the resulting quadratic must have exactly one solution (a repeated root). Missing this constraint can lead to accepting any value of B that makes \(\mathrm{x = -2}\) work, without verifying uniqueness.

2. Incorrect base conversion strategy

Some students get intimidated by the mixed bases (exponential on left, square root on right) and attempt complex logarithmic approaches instead of the simpler strategy of converting both sides to the same base. They might try taking log of both sides immediately, which creates unnecessary complexity and increases chances of computational errors.

Errors while executing the approach

1. Arithmetic errors with fractions

The problem involves multiple fractions (\(\frac{11}{2}\), \(\frac{3}{2}\)) and students frequently make mistakes when combining them. Common errors include: incorrectly converting 4 to \(\frac{8}{2}\), making sign errors when subtracting fractions (\(\frac{3}{2} - \frac{19}{2}\)), or miscalculating \(\frac{-16}{2} ÷ (-2)\). These fraction manipulations are where most computational errors occur.

2. Incorrect conversion of √8 to exponential form

Students often struggle with converting \(\sqrt{8} = \sqrt{2^3} = 2^{\frac{3}{2}}\). Common mistakes include writing \(\sqrt{8} = 2^{\frac{1}{2}}\) (forgetting that \(8 = 2^3\)), or incorrectly calculating the exponent as \(2^3\) or \(2^6\). This error propagates through the entire solution since it affects the target value for the exponent equation.

Errors while selecting the answer

1. Failing to verify the uniqueness condition

Even if students correctly calculate \(\mathrm{B = 4}\), many stop there without checking whether this actually gives a quadratic with exactly one solution. They might select \(\mathrm{B = 4}\) without confirming that \(\mathrm{x^2 + 4x + 4 = (x + 2)^2}\) indeed has only \(\mathrm{x = -2}\) as a solution. Some students might even get \(\mathrm{B = 4}\) but then doubt themselves because they didn't verify the uniqueness, leading them to second-guess and select a different answer choice.