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  • 15-03-2017
  • Mathematics
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Find two consecutive whole numbers such that the sum of their squares is 221.

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LammettHash
LammettHash LammettHash
  • 15-03-2017
[tex]n^2+(n+1)^2=2n^2+2n+1=221[/tex]
[tex]2n^2+2n=220[/tex]
[tex]n^2+n=110[/tex]
[tex]n^2+n-110=0[/tex]
[tex](n-10)(n+11)=0\implies n=10,n=-11[/tex]

But since [tex]n[/tex] must be a whole number, we ignore [tex]n=-11[/tex]. So the two integers are [tex]n=10[/tex] and [tex]n+1=11[/tex].
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