If (a+b√n) is the positive square root of (29-12√5), where a and b are integers, and n is a natural number, then the maximum possible value of (a+b+n) is
Question:
If (a+b√n) is the positive square root of (29-12√5), where a and b are integers, and n is a natural number, then the maximum possible value of (a+b+n) is
If (a+b√n) is the positive square root of (29-12√5), where a and b are integers, and n is a natural number, then the maximum possible value of (a+b+n) is
Options
Answer: 18
Explanation:
We are solving: √(29 − √125) = a + b√n, where a, b are integers and n is a natural number. Step 1: Square both sides (a + b√n)² = 29 − √125 → a² + b²n = 29 (the “normal numbers”) → 2ab√n = −√125 (the “square root part”) Step 2: Choose n to match the square root √125 = 5√5, so n should involve 5. After checking, n = 20 works with integers a = −3, b = 1. Check: a² + b²n = (−3)² + 1×20 = 9 + 20 = 29 ✅ 2ab√n = 2×(−3)×1×√20 = −6√20 ✅ (matches the square root part) Step 3: Find a + b + n a + b + n = −3 + 1 + 20 = 18 ✅ Answer: 18
Explanation:
We are solving: √(29 − √125) = a + b√n, where a, b are integers and n is a natural number. Step 1: Square both sides (a + b√n)² = 29 − √125 → a² + b²n = 29 (the “normal numbers”) → 2ab√n = −√125 (the “square root part”) Step 2: Choose n to match the square root √125 = 5√5, so n should involve 5. After checking, n = 20 works with integers a = −3, b = 1. Check: a² + b²n = (−3)² + 1×20 = 9 + 20 = 29 ✅ 2ab√n = 2×(−3)×1×√20 = −6√20 ✅ (matches the square root part) Step 3: Find a + b + n a + b + n = −3 + 1 + 20 = 18 ✅ Answer: 18
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