Difficulty: Medium
Correct Answer: 12
Explanation:
Introduction / Context:
This question tests a standard algebraic identity connecting x + y, xy, and x^2 + y^2. When you are given xy and x^2 + y^2, it is usually best not to solve for x and y individually. Instead, compute (x + y)^2 directly using (x + y)^2 = x^2 + y^2 + 2xy. That gives a single number, and then you take a square root. The “positive real numbers” condition matters because (x + y)^2 determines x + y only up to a sign, and positivity removes the negative possibility.
Given Data / Assumptions:
Concept / Approach:
Use the identity:
(x + y)^2 = x^2 + y^2 + 2xy.
Substitute the given values to find (x + y)^2, then take the square root. Because x and y are positive, x + y must be positive, so choose the positive root.
Step-by-Step Solution:
1) Write the identity:
(x + y)^2 = x^2 + y^2 + 2xy
2) Substitute x^2 + y^2 = 100:
(x + y)^2 = 100 + 2xy
3) Substitute xy = 22:
(x + y)^2 = 100 + 2*22 = 100 + 44
4) Simplify:
(x + y)^2 = 144
5) Take square root:
x + y = √144
6) Since x and y are positive, x + y is positive:
x + y = 12
Verification / Alternative check:
If x + y = 12 and xy = 22, then x and y are roots of t^2 - 12t + 22 = 0. The discriminant is 144 - 88 = 56, which is positive, so real solutions exist. Also, with sum 12 and product 22, both roots are positive (since product is positive and sum is positive). This is consistent with the assumption that x and y are positive real numbers.
Why Other Options Are Wrong:
• -12 is the negative root of √144, but x + y cannot be negative if x and y are positive.
• 6 is too small and would not give (x + y)^2 = 144.
• 72 and 144 are not values of x + y; 144 is (x + y)^2.
Common Pitfalls:
• Forgetting the identity includes +2xy, not -2xy.
• Computing (x + y)^2 correctly but then forgetting to take the square root.
• Ignoring the positivity condition and thinking both ±12 are valid.
Final Answer:
12
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