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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SHOW WORK SO I UNDERSTAND HOW TO DO IT THANK YOU
what work? If the smallest is x, then the next two are x+2 and x+4. So,
x^2 + (x+2)^2 = 54+20(x+4)
Now just crank it out.
To solve this problem, let's assume the first odd integer as x. Then, the next two consecutive odd integers will be x+2 and x+4.
According to the given information, the sum of squares of the first two integers is 54 more than 20 times the third integer. We can write this as an equation:
(x^2) + ((x+2)^2) = 20(x+4) + 54
Expanding and simplifying the equation:
x^2 + (x^2 + 4x +4) = 20x + 80 + 54
2x^2 + 4x + 4 = 20x + 134
Rearranging the equation to set it equal to zero:
2x^2 - 16x - 130 = 0
To solve this quadratic equation, we can either factor it or use the quadratic formula. Let's use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / 2a
In our equation, a=2, b=-16, and c=-130. Substituting these values into the quadratic formula:
x = (-(-16) ± √((-16)^2 - 4*2*(-130))) / (2*2)
x = (16 ± √(256 + 1040)) / 4
x = (16 ± √(1296)) / 4
x = (16 ± 36) / 4
This gives us two possible solutions for x:
x1 = (16 + 36) / 4 = 52 / 4 = 13
x2 = (16 - 36) / 4 = -20 / 4 = -5
Now, we can find the remaining two consecutive odd integers, x+2 and x+4:
For x1 = 13:
- The first integer = 13
- The second integer = 13 + 2 = 15
- The third integer = 13 + 4 = 17
For x2 = -5:
- The first integer = -5
- The second integer = -5 + 2 = -3
- The third integer = -5 + 4 = -1
Therefore, the two sets of three consecutive odd integers that satisfy the given conditions are:
1) 13, 15, 17
2) -5, -3, -1