Three consecutive odd intergers are such that the sum of the squares of the first two integers is 54 more than 20 times the third integer. Determine the three integers.
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To solve this problem, let's start by assigning variables to the three consecutive odd integers. Let's call the first odd integer "x", the second odd integer "x + 2", and the third odd integer "x + 4".
Now let's set up the equation based on the given information: "the sum of the squares of the first two integers is 54 more than 20 times the third integer."
Accordingly, we have:
x^2 + (x + 2)^2 = 20(x + 4) + 54
To solve the equation, we need to simplify it step by step.
Expanding the square terms:
x^2 + (x^2 + 4x + 4) = 20x + 80 + 54
Simplifying the equation:
2x^2 + 4x + 4 = 20x + 134
Next, let's move all terms to one side of the equation by subtracting 20x and 134 from both sides:
2x^2 + 4x + 4 - 20x - 134 = 0
Simplifying further:
2x^2 - 16x - 130 = 0
Now, we can solve this quadratic equation by factoring or by using the quadratic formula. Let's use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this equation, a = 2, b = -16, and c = -130. Substituting these values into the quadratic formula:
x = (-(-16) ± √((-16)^2 - 4 * 2 * -130)) / (2 * 2)
Simplifying inside the square root:
x = (16 ± √(256 + 1040)) / 4
x = (16 ± √(1296)) / 4
x = (16 ± 36) / 4
x = (16 + 36) / 4 or x = (16 - 36) / 4
Simplifying further:
x = 52 / 4 or x = -20 / 4
x = 13 or x = -5
Since we are dealing with odd integers, we can discard the -5 solution, leaving us with x = 13.
Therefore, the three consecutive odd integers are:
1st integer: x = 13
2nd integer: x + 2 = 13 + 2 = 15
3rd integer: x + 4 = 13 + 4 = 17
So, the three consecutive odd integers are 13, 15, and 17.