SOLUTION: here is the problem : Find three consecutive integers such that the sum of the squares of the second and third exceeds the square of the first by 77. please!!thanks.

Question 235405: here is the problem : Find three consecutive integers such that the sum of the squares of the second and third exceeds the square of the first by 77. please!!thanks.
Answer by checkley77(12844) (Show Source): You can put this solution on YOUR website!
Let x, x+1 & x+2 be the 3 integers.
(x+1)^2+(x+2)^2=x^2+77
x^2+2x+1+x^2+4x+4=x^2+77
2x^2-x^2+6x+5-77=0
x^2+6x-72=0
(x+12)(x-6)=0
x+12=0
x=-12 ans.
x-6=0
x=6 ans.
Proofs:
(-12+1)^2+-12+2)^2=-12^2+77
-11^2+-10^2=144+77
121+100=221
221=221
(6+1)^2+(6+2)^2=6^2+77
7^2+8^2=36+77
49+64=113
113=113
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