SOLUTION: find 2 consecutive integers whose sum is 16 less than the square of the larger number.
i'm not sure if i set it up right.
this is how i set it up : x+x+1={{{ sqrt (x+1) }}}
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-> SOLUTION: find 2 consecutive integers whose sum is 16 less than the square of the larger number.
i'm not sure if i set it up right.
this is how i set it up : x+x+1={{{ sqrt (x+1) }}}
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Question 335481: find 2 consecutive integers whose sum is 16 less than the square of the larger number.
i'm not sure if i set it up right.
this is how i set it up : x+x+1= -16 Answer by mananth(16946) (Show Source):
You can put this solution on YOUR website! let the two cosequtive integers be x & x+1
..
x+x+1 = (x+1)^2-16
2x+1=x^2+2x+1-16
add -2x to both sides
2x+1-2x=x^2+2x-2x-16+1
1=x^2+1-16
add -1
x^2-16=0
(x+4)(x-4)=0
x=-4 or +4
The integers are -4,-5 OR 4,5
