Asked by Victory
Find 2 consecutive odd number whose product is 224
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GPT 3.5
Let's assume the two consecutive odd numbers are n and n + 2.
We are given that the product of these two numbers is 224. Therefore, we can set up the following equation:
n(n + 2) = 224
Expanding the equation, we get:
n^2 + 2n = 224
Rearranging the equation and simplifying, we have:
n^2 + 2n - 224 = 0
To solve this quadratic equation, we can either factor it or use the quadratic formula. In this case, factoring is not feasible, so we will use the quadratic formula:
n = (-b ± √(b^2 - 4ac)) / (2a)
For this equation, a = 1, b = 2, and c = -224. Substituting these values into the quadratic formula, we can solve for n.
n = (-2 ± √(2^2 - 4(1)(-224))) / (2(1))
n = (-2 ± √(4 + 896)) / 2
n = (-2 ± √900) / 2
n = (-2 ± 30) / 2
Now we have two possible values for n:
n = (-2 + 30) / 2 = 28 / 2 = 14
n = (-2 - 30) / 2 = -32 / 2 = -16
Therefore, the two consecutive odd numbers whose product is 224 are 14 and 16.
We are given that the product of these two numbers is 224. Therefore, we can set up the following equation:
n(n + 2) = 224
Expanding the equation, we get:
n^2 + 2n = 224
Rearranging the equation and simplifying, we have:
n^2 + 2n - 224 = 0
To solve this quadratic equation, we can either factor it or use the quadratic formula. In this case, factoring is not feasible, so we will use the quadratic formula:
n = (-b ± √(b^2 - 4ac)) / (2a)
For this equation, a = 1, b = 2, and c = -224. Substituting these values into the quadratic formula, we can solve for n.
n = (-2 ± √(2^2 - 4(1)(-224))) / (2(1))
n = (-2 ± √(4 + 896)) / 2
n = (-2 ± √900) / 2
n = (-2 ± 30) / 2
Now we have two possible values for n:
n = (-2 + 30) / 2 = 28 / 2 = 14
n = (-2 - 30) / 2 = -32 / 2 = -16
Therefore, the two consecutive odd numbers whose product is 224 are 14 and 16.
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