How many values of x are there for which \(\frac{1}{\mathrm{x}(\mathrm{x} + 1)} + \frac{1}{\mathrm{x}(\mathrm{x} + 2)} = \frac{1}{(\mathrm{x} + 1)}\)

1. Translate the problem requirements: Find how many values of x satisfy the equation where we have fractions with x in denominators, meaning we need values that make the equation true but don't make any denominator zero
2. Identify domain restrictions: Determine which x values are forbidden because they make denominators zero
3. Simplify the equation algebraically: Combine fractions and manipulate the equation to isolate terms and reduce complexity
4. Solve the resulting equation: Work with the simplified form to find potential x values
5. Verify solutions are valid: Check that our solutions don't violate domain restrictions and actually satisfy the original equation

Execution of Strategic Approach

1. Translate the problem requirements

We need to find how many values of x make this equation true: \(\frac{1}{\mathrm{x}(\mathrm{x} + 1)} + \frac{1}{\mathrm{x}(\mathrm{x} + 2)} = \frac{1}{(\mathrm{x} + 1)}\)

In plain English, we're looking for numbers that we can substitute for x that will make both sides of the equation equal. However, we have to be careful - since x appears in the bottom of fractions, we can't use any value of x that would make a denominator zero (because division by zero is undefined).

Process Skill: TRANSLATE - Converting the equation into a clear understanding of what we're seeking

2. Identify domain restrictions

Let's think about which values of x we absolutely cannot use:

• If \(\mathrm{x} = 0\), then \(\mathrm{x}(\mathrm{x} + 1) = 0\) and \(\mathrm{x}(\mathrm{x} + 2) = 0\), making our fractions undefined
• If \(\mathrm{x} = -1\), then \(\mathrm{x}(\mathrm{x} + 1) = 0\) and \((\mathrm{x} + 1) = 0\), making our fractions undefined
• If \(\mathrm{x} = -2\), then \(\mathrm{x}(\mathrm{x} + 2) = 0\), making our second fraction undefined

So we must exclude \(\mathrm{x} = 0\), \(\mathrm{x} = -1\), and \(\mathrm{x} = -2\) from any potential solutions.

Process Skill: APPLY CONSTRAINTS - Identifying forbidden values before solving

3. Simplify the equation algebraically

Now let's work with the equation step by step. We'll find a common way to combine the left side.

Starting with: \(\frac{1}{\mathrm{x}(\mathrm{x} + 1)} + \frac{1}{\mathrm{x}(\mathrm{x} + 2)} = \frac{1}{(\mathrm{x} + 1)}\)

To add the fractions on the left, we need a common denominator. The common denominator for \(\mathrm{x}(\mathrm{x} + 1)\) and \(\mathrm{x}(\mathrm{x} + 2)\) is \(\mathrm{x}(\mathrm{x} + 1)(\mathrm{x} + 2)\).

\(\frac{1}{\mathrm{x}(\mathrm{x} + 1)} = \frac{\mathrm{x} + 2}{\mathrm{x}(\mathrm{x} + 1)(\mathrm{x} + 2)}\)

\(\frac{1}{\mathrm{x}(\mathrm{x} + 2)} = \frac{\mathrm{x} + 1}{\mathrm{x}(\mathrm{x} + 1)(\mathrm{x} + 2)}\)

So our equation becomes:

\(\frac{\mathrm{x} + 2}{\mathrm{x}(\mathrm{x} + 1)(\mathrm{x} + 2)} + \frac{\mathrm{x} + 1}{\mathrm{x}(\mathrm{x} + 1)(\mathrm{x} + 2)} = \frac{1}{\mathrm{x} + 1}\)

Combining the left side:

\(\frac{(\mathrm{x} + 2) + (\mathrm{x} + 1)}{\mathrm{x}(\mathrm{x} + 1)(\mathrm{x} + 2)} = \frac{1}{\mathrm{x} + 1}\)

\(\frac{2\mathrm{x} + 3}{\mathrm{x}(\mathrm{x} + 1)(\mathrm{x} + 2)} = \frac{1}{\mathrm{x} + 1}\)

4. Solve the resulting equation

Now we have: \(\frac{2\mathrm{x} + 3}{\mathrm{x}(\mathrm{x} + 1)(\mathrm{x} + 2)} = \frac{1}{\mathrm{x} + 1}\)

Since we know \(\mathrm{x} \neq -1\), we can multiply both sides by \((\mathrm{x} + 1)\):

\(\frac{2\mathrm{x} + 3}{\mathrm{x}(\mathrm{x} + 2)} = 1\)

This means: \(2\mathrm{x} + 3 = \mathrm{x}(\mathrm{x} + 2)\)

Expanding the right side: \(2\mathrm{x} + 3 = \mathrm{x}^2 + 2\mathrm{x}\)

Subtracting \(2\mathrm{x}\) from both sides: \(3 = \mathrm{x}^2\)

Therefore: \(\mathrm{x}^2 = 3\), which gives us \(\mathrm{x} = \sqrt{3}\) or \(\mathrm{x} = -\sqrt{3}\)

Process Skill: MANIPULATE - Systematic algebraic manipulation to isolate the variable

5. Verify solutions are valid

We found two potential solutions: \(\mathrm{x} = \sqrt{3}\) and \(\mathrm{x} = -\sqrt{3}\)

First, let's check that neither violates our domain restrictions:

• \(\sqrt{3} \approx 1.73\), which is not 0, -1, or -2 ✓
• \(-\sqrt{3} \approx -1.73\), which is not 0, -1, or -2 ✓

Now let's verify these actually work in the original equation. For \(\mathrm{x} = \sqrt{3}\):

Left side: \(\frac{1}{\sqrt{3}(\sqrt{3} + 1)} + \frac{1}{\sqrt{3}(\sqrt{3} + 2)} = \frac{1}{\sqrt{3}(\sqrt{3} + 1)} + \frac{1}{\sqrt{3}(\sqrt{3} + 2)}\)

Right side: \(\frac{1}{\sqrt{3} + 1}\)

Since our algebraic work was valid and we derived these solutions correctly, both \(\mathrm{x} = \sqrt{3}\) and \(\mathrm{x} = -\sqrt{3}\) are valid solutions.

4. Final Answer

We found exactly two values of x that satisfy the equation: \(\mathrm{x} = \sqrt{3}\) and \(\mathrm{x} = -\sqrt{3}\).

Therefore, the answer is C. Two.

Common Faltering Points

Errors while devising the approach

1. Missing or incorrectly identifying domain restrictions

Students often jump straight into solving the equation without first identifying which values of x make the denominators zero. They might miss that \(\mathrm{x} = 0\), \(\mathrm{x} = -1\), and \(\mathrm{x} = -2\) are forbidden values, which is crucial because any solution that equals one of these values must be rejected.

2. Misunderstanding what the question is asking

Some students might confuse "how many values of x" with "what are the values of x" and spend time trying to find the exact decimal values rather than simply counting the number of solutions.

Errors while executing the approach

1. Common denominator errors when combining fractions

When finding the common denominator for \(\frac{1}{\mathrm{x}(\mathrm{x} + 1)} + \frac{1}{\mathrm{x}(\mathrm{x} + 2)}\), students often make mistakes in the algebraic manipulation. They might incorrectly write the equivalent fractions or make errors when combining the numerators \((\mathrm{x} + 2) + (\mathrm{x} + 1) = 2\mathrm{x} + 3\).

2. Sign errors when solving the quadratic

After reaching \(\mathrm{x}^2 = 3\), students frequently forget that this equation has two solutions: both \(\mathrm{x} = \sqrt{3}\) and \(\mathrm{x} = -\sqrt{3}\). They might only consider the positive square root and miss the negative solution.

3. Algebraic manipulation errors

When cross-multiplying or expanding \(\mathrm{x}(\mathrm{x} + 2) = \mathrm{x}^2 + 2\mathrm{x}\), students can make basic algebraic errors that lead to incorrect equations and wrong final answers.

Errors while selecting the answer

1. Forgetting to verify solutions against domain restrictions

Even after finding \(\mathrm{x} = \sqrt{3}\) and \(\mathrm{x} = -\sqrt{3}\), students might forget to check whether these values violate the domain restrictions they identified earlier. However, in this case, both solutions are valid since neither equals 0, -1, or -2.

2. Counting solutions incorrectly

Students might find both solutions but then incorrectly count them, perhaps thinking they found one solution with two forms rather than recognizing these as two distinct values of x.