Question 757773: find three consecutive odd integers such that 10 times the first when added to the third is equal to the square of the second
Answer by sachi(548) (Show Source):
You can put this solution on YOUR website! let the 3 consecutive odd integers are x-2,x,x+2
10 times the first when added to the third is equal to the square of the second
so 10(x-2)+(x+2)=x^2
or 10x-20+x+2=x^2
or x^2-11x+18=0 by simplifying
or x^2 -2x-9x+18=0
or x(x-2)-9(x-2)=o
or (x-2)(x-9)=0
or x=2 or 9 =9 rejecting the even no 2
so the 3 consecutive odd integers are either 7,9,11
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