Asked by LP
                Find three consecutive odd integers such that the square of the first increased by product of the other two is 224.
            
            
        Answers
                    Answered by
            Reiny
            
    let the middle number be x
then the 1st is x-2
and the 3rd is x+2
(x-2)^2 + x(x+2) = 224
x^2 - 4x + 4 + x^2 + 2x = 224
2x^2 -2x - 220 = 0
x^2 - x - 110 = 0
(x-11)(x+10) = 0
x = 11 or x = -10, but x is to be odd
so
x = 11
and the three numbers are 9, 11, and 13
check:
9^2 + 11(13) = 224 , YEAH!
    
then the 1st is x-2
and the 3rd is x+2
(x-2)^2 + x(x+2) = 224
x^2 - 4x + 4 + x^2 + 2x = 224
2x^2 -2x - 220 = 0
x^2 - x - 110 = 0
(x-11)(x+10) = 0
x = 11 or x = -10, but x is to be odd
so
x = 11
and the three numbers are 9, 11, and 13
check:
9^2 + 11(13) = 224 , YEAH!
                    Answered by
            gricelis
            
    i got this so correct omg im so happy
    
                                                    There are no AI answers yet. The ability to request AI answers is coming soon!
                                            
                Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.