- Translate the problem requirements: We need to find the value of a fraction with exponential expressions in both numerator and denominator, where the exponents are \((x+y)^2\) and \((x-y)^2\) respectively, and we're given that \(xy = 1\).
- Apply exponent rules to simplify the fraction: Use the property that when dividing powers with the same base, we subtract exponents to get \(2^{[(x+y)^2 - (x-y)^2]}\).
- Expand and simplify the exponent expression: Expand both squared terms and combine like terms to find the simplified exponent.
- Use the given constraint to evaluate: Apply \(xy = 1\) to determine the final numerical value.
Execution of Strategic Approach
1. Translate the problem requirements
Let's break down what we're looking for in plain English. We have a fraction where both the top and bottom are powers of 2. The top has 2 raised to the power of \((x+y)^2\), and the bottom has 2 raised to the power of \((x-y)^2\). We need to find the actual numerical value of this entire expression, and we're told that \(xy = 1\).
In mathematical terms, we want to find the value of: \(\frac{2^{(x+y)^2}}{2^{(x-y)^2}}\)
Process Skill: TRANSLATE - Converting the complex exponential expression into a clear mathematical goal
2. Apply exponent rules to simplify the fraction
When we have the same base (in this case, 2) in both the numerator and denominator, we can use a fundamental rule: dividing powers means subtracting the exponents.
Think of it this way: if you have \(2^3 \div 2^1\), that's the same as \(2^{3-1} = 2^2\). The same principle applies here, even though our exponents are more complex.
So our fraction becomes: \(2^{[(x+y)^2 - (x-y)^2]}\)
Now we need to figure out what \((x+y)^2 - (x-y)^2\) equals.
3. Expand and simplify the exponent expression
Let's expand each squared term step by step using the basic algebra pattern \((a+b)^2 = a^2 + 2ab + b^2\):
\((x+y)^2 = x^2 + 2xy + y^2\)
\((x-y)^2 = x^2 - 2xy + y^2\)
Now we subtract the second from the first:
\((x+y)^2 - (x-y)^2 = (x^2 + 2xy + y^2) - (x^2 - 2xy + y^2)\)
When we distribute the negative sign and combine like terms:
\(= x^2 + 2xy + y^2 - x^2 + 2xy - y^2\)
\(= (x^2 - x^2) + (y^2 - y^2) + (2xy + 2xy)\)
\(= 0 + 0 + 4xy\)
\(= 4xy\)
So our expression becomes: \(2^{4xy}\)
Process Skill: MANIPULATE - Using algebraic expansion and combining like terms to simplify
4. Use the given constraint to evaluate
Now we can use the fact that \(xy = 1\). Since we found that our exponent is \(4xy\), we can substitute:
\(2^{4xy} = 2^{4 \times 1} = 2^4\)
Calculating \(2^4\):
\(2^4 = 2 \times 2 \times 2 \times 2 = 16\)
Process Skill: APPLY CONSTRAINTS - Using the given condition xy = 1 to reach the final numerical answer
Final Answer
The value of \(\frac{2^{(x+y)^2}}{2^{(x-y)^2}}\) is 16.
This matches answer choice D, confirming our solution is correct.
Common Faltering Points
Errors while devising the approach
- Misunderstanding the constraint xy = 1: Students may think they need to find specific values of x and y (like x = 1, y = 1) rather than recognizing that the product \(xy = 1\) is sufficient to solve the problem. This leads them down unnecessary paths of trying to solve for individual variables.
- Not recognizing the exponent rule application: Students might attempt to evaluate \(2^{(x+y)^2}\) and \(2^{(x-y)^2}\) separately instead of recognizing they can use the quotient rule for exponents (\(a^m \div a^n = a^{(m-n)}\)) to simplify the fraction first.
- Overlooking algebraic expansion strategy: Students may not realize they need to expand \((x+y)^2 - (x-y)^2\) algebraically and might try to substitute values or use other complex approaches unnecessarily.
Errors while executing the approach
- Algebraic expansion errors: When expanding \((x+y)^2\) and \((x-y)^2\), students commonly make sign errors, especially with \((x-y)^2 = x^2 - 2xy + y^2\). They might write it as \(x^2 + 2xy + y^2\) or forget the middle term entirely.
- Combining like terms incorrectly: When subtracting \((x+y)^2 - (x-y)^2\), students often make errors distributing the negative sign or combining terms. They might get \(2xy\) instead of \(4xy\), or make other arithmetic mistakes in the simplification.
- Substitution timing errors: Students might try to substitute \(xy = 1\) too early in the process, before properly simplifying the algebraic expression, leading to confusion and incorrect calculations.
Errors while selecting the answer
- Calculation errors with powers of 2: Even after correctly arriving at \(2^4\), students might miscalculate this basic power. Common errors include \(2^4 = 8\) (confusing it with \(2^3\)) or \(2^4 = 32\) (confusing it with \(2^5\)).
Alternate Solutions
Smart Numbers Approach
Step 1: Choose specific values that satisfy xy = 1
Let's select \(x = 2\) and \(y = 1/2\), since \(2 \times (1/2) = 1\). These values are chosen because:
- They satisfy our constraint \(xy = 1\)
- They give us nice, workable numbers for \(x + y\) and \(x - y\)
- The calculations will be straightforward
Step 2: Calculate the sum and difference
- \(x + y = 2 + 1/2 = 5/2\)
- \(x - y = 2 - 1/2 = 3/2\)
Step 3: Calculate the squared terms
- \((x + y)^2 = (5/2)^2 = 25/4\)
- \((x - y)^2 = (3/2)^2 = 9/4\)
Step 4: Substitute into the original expression
\(\frac{2^{(x+y)^2}}{2^{(x-y)^2}} = \frac{2^{25/4}}{2^{9/4}}\)
Step 5: Apply exponent rules
\(\frac{2^{25/4}}{2^{9/4}} = 2^{25/4 - 9/4} = 2^{16/4} = 2^4 = 16\)
Verification with different smart numbers:
Let's verify with \(x = 1/2\) and \(y = 2\):
- \(x + y = 1/2 + 2 = 5/2\)
- \(x - y = 1/2 - 2 = -3/2\)
- \((x + y)^2 = 25/4\)
- \((x - y)^2 = (-3/2)^2 = 9/4\)
This gives us the same result: \(2^{25/4 - 9/4} = 2^4 = 16\)
The answer is D: 16.
