Find 3 consecutive numbers where the product of the smaller two numbers is 19 less than the square of the largest number.
1 answer:
N; n+1; n+2 - 3 consecutive numbers
n(n + 1) = (n + 2)² - 19 |use a(b + c) = ab + ac and (a + b)² = a² + 2ab + b²
n² + n = n² + 4n + 4 - 19 |subtract n² from both sides
n = 4n - 15 |subtract 4n from both sides
-3n = -15 |divide both sides by (-3)
n = 5
n + 1 = 5 + 1 = 6
n + 2 = 5 + 2 = 7
Answer: 5; 6; 7.
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Step-by-step explanation: