SOLUTION: Find three consecutive odd integers such that the sum of the first and the square of the last is 236.

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Question 409110: Find three consecutive odd integers such that the sum of the first and the square of the last is 236.
Answer by Adrish(44) About Me  (Show Source):
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SOLUTIONS: Let the three consecutive odd integers are x,(x+2),(x+4)
Given:x+(x+4)^2 = 236
x+ x^2 +8x +16 = 236
x^2 +9x - 220 =0
By solving the quadratic equation
x^2 + 20 x -11x -220 =0
x(x+20) -11(x+20)=0
(x+20)(x-11)=0
x+20=0
and x-11=0
so x=-20 and x = 11
so the three consecutive integers are 11. 11+2, 11+4
11,13,15