How do you find the three consecutive integers such that the sum of the squares of the second and the third exceeds the square of the first by 21?

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Answer 1 · 2016-03-08T16:18:40

-8,-7,-6 or 2,3,4

Three consecutive numbers

$n-1,n, n+1$

The sum of the squares of the second and the third exceeds the square of the first by 21:

$n^2+(n+1)^2=(n-1)^2+21$

$n^2+(n^2+2n+1)=(n^2-2n+1)+21$

$n^2+4n-21=0$

$n=(-4+-sqrt(4^2-4*1*(-21)))/2$

$n=(-4+-sqrt(100))/2$

$n=(-4+-10)/2$

$n=-14/2=-7 or n=6/2=3$

So you have the solutions:

-8,-7,-6 and 2,3,4