SOLUTION: I was given a question to answer, but i do not understand it. Here is the question:
When the square of the third consecutive odd integer is added to three times the second integer
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-> SOLUTION: I was given a question to answer, but i do not understand it. Here is the question:
When the square of the third consecutive odd integer is added to three times the second integer
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Question 983089: I was given a question to answer, but i do not understand it. Here is the question:
When the square of the third consecutive odd integer is added to three times the second integer, the result is 92 more than twice the first. Find the integer.
Thank you for helping and God Bless! Found 2 solutions by Alan3354, macston:Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website! .
When the square of the third consecutive odd integer is added to three times the second integer, the result is 92 more than twice the first. Find the integer.
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We must have three consecutive odd integers:
Let x, x+2, x+4 be 3 consecutive odd integers.
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The square of the third: (x+4)^2
Added to 3 times second:(x+4)^2+3(x+2)
Result is 92 more than twice first:
(92 more than twice first=2x+92)
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(x+4)^2+3(x+2)=2x+92
(x^2+8x+16)+(3x+6)-(2x+92)=0
x^2+8x+16+3x+6-2x-92=0
x^2+9x-70=0
(x+14)(x-5)=0
x+14=0 OR x-5=0
x=-14 OR x=5
-14 is not an odd integer, so the first odd integer is 5.
Second integer=x+2=5+2=7
Third integer=x+4=5+4=9
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The three integers are 5,7, and 9.
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CHECK:
Square of third added to 3 times second is twice first plus 92:
.
9^2+3(7)=2(5)+92
81+21=102
102=102
