If c is a nonnegative constant, how many real number values for x satisfy the equation sqrt(c - x) =...

1. Translate the problem requirements: We need to find how many real number values of x satisfy \(\sqrt{\mathrm{c} - \mathrm{x}} = \sqrt{\mathrm{c}} - \sqrt{\mathrm{x}}\), where c is nonnegative. This means finding values where the square root of (c minus x) equals the difference between the square root of c and the square root of x.
2. Establish domain constraints: Determine what values of x are mathematically valid for both sides of the equation to exist as real numbers.
3. Analyze the equation structure: Square both sides to eliminate square roots and reveal the underlying algebraic relationship.
4. Solve for potential solutions: Use algebraic manipulation to find candidate values of x that satisfy the equation.
5. Verify solutions within domain: Check that our solutions satisfy both the original equation and the domain constraints to count valid solutions.

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what this equation is asking. We have \(\sqrt{\mathrm{c} - \mathrm{x}} = \sqrt{\mathrm{c}} - \sqrt{\mathrm{x}}\), where c is nonnegative.

In everyday language: We're looking for values of x where "the square root of (c minus x)" equals "the square root of c minus the square root of x".

Think of it this way: if c = 9 and x = 4, then the left side would be \(\sqrt{9-4} = \sqrt{5}\), and the right side would be \(\sqrt{9} - \sqrt{4} = 3 - 2 = 1\). Since \(\sqrt{5} ≠ 1\), x = 4 wouldn't work when c = 9.

Process Skill: TRANSLATE - Converting the equation language into clear mathematical understanding

2. Establish domain constraints

Before we solve, we need to figure out what values of x are even allowed.

For the left side \(\sqrt{\mathrm{c} - \mathrm{x}}\) to be real, we need: \(\mathrm{c} - \mathrm{x} ≥ 0\), which means \(\mathrm{x} ≤ \mathrm{c}\).

For the right side \(\sqrt{\mathrm{c}} - \sqrt{\mathrm{x}}\) to be real, we need: \(\sqrt{\mathrm{x}}\) to exist, which means \(\mathrm{x} ≥ 0\).

So our domain is: \(0 ≤ \mathrm{x} ≤ \mathrm{c}\)

Also, for the right side to be nonnegative (since the left side is always nonnegative), we need \(\sqrt{\mathrm{c}} - \sqrt{\mathrm{x}} ≥ 0\), which means \(\sqrt{\mathrm{c}} ≥ \sqrt{\mathrm{x}}\), so \(\mathrm{x} ≤ \mathrm{c}\) (which we already have).

Process Skill: APPLY CONSTRAINTS - Determining valid input ranges before solving

3. Analyze the equation structure

Now let's work with the equation \(\sqrt{\mathrm{c} - \mathrm{x}} = \sqrt{\mathrm{c}} - \sqrt{\mathrm{x}}\).

To eliminate the square roots, let's square both sides:
\((\sqrt{\mathrm{c} - \mathrm{x}})^2 = (\sqrt{\mathrm{c}} - \sqrt{\mathrm{x}})^2\)

Left side: \(\mathrm{c} - \mathrm{x}\)

Right side: \((\sqrt{\mathrm{c}} - \sqrt{\mathrm{x}})^2 = (\sqrt{\mathrm{c}})^2 - 2\sqrt{\mathrm{c}}\sqrt{\mathrm{x}} + (\sqrt{\mathrm{x}})^2 = \mathrm{c} - 2\sqrt{\mathrm{cx}} + \mathrm{x}\)

So our equation becomes:
\(\mathrm{c} - \mathrm{x} = \mathrm{c} - 2\sqrt{\mathrm{cx}} + \mathrm{x}\)

4. Solve for potential solutions

Simplifying the equation \(\mathrm{c} - \mathrm{x} = \mathrm{c} - 2\sqrt{\mathrm{cx}} + \mathrm{x}\):

Subtract c from both sides: \(-\mathrm{x} = -2\sqrt{\mathrm{cx}} + \mathrm{x}\)

Subtract x from both sides: \(-2\mathrm{x} = -2\sqrt{\mathrm{cx}}\)

Divide by -2: \(\mathrm{x} = \sqrt{\mathrm{cx}}\)

Now square both sides to eliminate the remaining square root:
\(\mathrm{x}^2 = \mathrm{cx}\)

Factor: \(\mathrm{x}^2 - \mathrm{cx} = 0\), so \(\mathrm{x}(\mathrm{x} - \mathrm{c}) = 0\)

This gives us: \(\mathrm{x} = 0\) or \(\mathrm{x} = \mathrm{c}\)

5. Verify solutions within domain

We found \(\mathrm{x} = 0\) and \(\mathrm{x} = \mathrm{c}\) as potential solutions. Let's check if they work in the original equation:

Check x = 0:
Left side: \(\sqrt{\mathrm{c} - 0} = \sqrt{\mathrm{c}}\)
Right side: \(\sqrt{\mathrm{c}} - \sqrt{0} = \sqrt{\mathrm{c}} - 0 = \sqrt{\mathrm{c}}\)
Since \(\sqrt{\mathrm{c}} = \sqrt{\mathrm{c}}\) ✓, x = 0 works!

Check x = c:
Left side: \(\sqrt{\mathrm{c} - \mathrm{c}} = \sqrt{0} = 0\)
Right side: \(\sqrt{\mathrm{c}} - \sqrt{\mathrm{c}} = 0\)
Since \(0 = 0\) ✓, x = c works!

Both solutions are in our domain \(0 ≤ \mathrm{x} ≤ \mathrm{c}\), and both satisfy the original equation.

Process Skill: CONSIDER ALL CASES - Verifying each potential solution in the original equation

4. Final Answer

We found exactly 2 real number values for x that satisfy the equation: \(\mathrm{x} = 0\) and \(\mathrm{x} = \mathrm{c}\).

The answer is C: 2

Common Faltering Points

Errors while devising the approach

1. Overlooking Domain Constraints

Many students jump straight into algebraic manipulation without first establishing what values of x are mathematically valid. They miss that \(\sqrt{\mathrm{x}}\) requires \(\mathrm{x} ≥ 0\) and \(\sqrt{\mathrm{c}-\mathrm{x}}\) requires \(\mathrm{x} ≤ \mathrm{c}\), leading to invalid solutions or confusion about which solutions to accept.

2. Misunderstanding the Right Side Structure

Students often misinterpret \(\sqrt{\mathrm{c}} - \sqrt{\mathrm{x}}\) as \(\sqrt{\mathrm{c}-\mathrm{x}}\), treating it as one square root expression instead of recognizing it as the difference of two separate square root terms. This fundamental misreading leads to completely different equations.

3. Forgetting the Non-negativity Requirement

Since the left side \(\sqrt{\mathrm{c}-\mathrm{x}}\) is always non-negative, students may forget that the right side \(\sqrt{\mathrm{c}} - \sqrt{\mathrm{x}}\) must also be non-negative. This means \(\sqrt{\mathrm{c}} ≥ \sqrt{\mathrm{x}}\), which provides an additional constraint that some students overlook when setting up their approach.

Errors while executing the approach

1. Expanding the Squared Right Side Incorrectly

When squaring \((\sqrt{\mathrm{c}} - \sqrt{\mathrm{x}})^2\), students frequently make the algebra error of writing it as \(\mathrm{c} - \mathrm{x}\) instead of the correct expansion \(\mathrm{c} - 2\sqrt{\mathrm{cx}} + \mathrm{x}\). They forget the middle cross-term \(-2\sqrt{\mathrm{c}}\sqrt{\mathrm{x}}\), leading to an incorrect simplified equation.

2. Sign Errors During Algebraic Manipulation

In the step where \(-\mathrm{x} = -2\sqrt{\mathrm{cx}} + \mathrm{x}\) becomes \(-2\mathrm{x} = -2\sqrt{\mathrm{cx}}\), students often lose track of negative signs. Some may incorrectly write \(2\mathrm{x} = 2\sqrt{\mathrm{cx}}\) or make other sign errors that lead to wrong solutions.

3. Introducing Extraneous Solutions

When squaring both sides of equations (done twice in this problem), students may not realize they could be introducing solutions that don't satisfy the original equation. They solve \(\mathrm{x}(\mathrm{x}-\mathrm{c}) = 0\) to get \(\mathrm{x} = 0\) and \(\mathrm{x} = \mathrm{c}\) but fail to check these back in the original equation.

Errors while selecting the answer

1. Counting Invalid Solutions

Students might find potential solutions like \(\mathrm{x} = 0\) and \(\mathrm{x} = \mathrm{c}\) but fail to verify them in the original equation. If they don't check their work, they might count solutions that are actually extraneous, leading to an incorrect count.

2. Misunderstanding 'How Many Values'

Some students might think the question asks for the actual values of x rather than how many such values exist. They might try to select an answer choice that lists the solutions (0 and c) rather than recognizing that '2' represents the count of solutions.

Alternate Solutions

Smart Numbers Approach

Instead of working algebraically with the general case, we can test specific values of c to identify the pattern of solutions.

Step 1: Choose a convenient value for c

Let's use c = 4 (a perfect square to make square root calculations clean).
The equation becomes: \(\sqrt{4 - \mathrm{x}} = \sqrt{4} - \sqrt{\mathrm{x}} = 2 - \sqrt{\mathrm{x}}\)

Step 2: Determine the domain

For real solutions, we need:
• \(4 - \mathrm{x} ≥ 0\), so \(\mathrm{x} ≤ 4\)
• \(\mathrm{x} ≥ 0\) (for \(\sqrt{\mathrm{x}}\) to be real)
Therefore: \(0 ≤ \mathrm{x} ≤ 4\)

Step 3: Solve the equation

\(\sqrt{4 - \mathrm{x}} = 2 - \sqrt{\mathrm{x}}\)
Squaring both sides: \(4 - \mathrm{x} = (2 - \sqrt{\mathrm{x}})^2\)
\(4 - \mathrm{x} = 4 - 4\sqrt{\mathrm{x}} + \mathrm{x}\)
\(-\mathrm{x} = -4\sqrt{\mathrm{x}} + \mathrm{x}\)
\(2\mathrm{x} = 4\sqrt{\mathrm{x}}\)
\(\mathrm{x} = 2\sqrt{\mathrm{x}}\)

Step 4: Find solutions

Case 1: If \(\mathrm{x} = 0\), then \(0 = 2\sqrt{0} = 0\) ✓
Case 2: If \(\mathrm{x} > 0\), divide by \(\sqrt{\mathrm{x}}\): \(\sqrt{\mathrm{x}} = 2\), so \(\mathrm{x} = 4\)

Step 5: Verify solutions

For \(\mathrm{x} = 0\): \(\sqrt{4-0} = 2\) and \(\sqrt{4} - \sqrt{0} = 2 - 0 = 2\) ✓
For \(\mathrm{x} = 4\): \(\sqrt{4-4} = 0\) and \(\sqrt{4} - \sqrt{4} = 2 - 2 = 0\) ✓

Step 6: Confirm with another value

Let's verify with c = 9:
Following the same process, we get \(\mathrm{x} = 0\) and \(\mathrm{x} = 9\) as solutions.

Step 7: Identify the pattern

For any positive value of c, we consistently get exactly 2 solutions: \(\mathrm{x} = 0\) and \(\mathrm{x} = \mathrm{c}\).
Therefore, the answer is 2.