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The positive difference between two consecutive even perfect squares is 268 . Compute the larger of the two squares.
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so it is (x+1)^2-x^2=268
says even right, so maybe \[(2n+2)^2-(2n)^2\]
x^2+1^2-x^2=268 x^2+1-x^2=268 x^2-x^2=268-1 0=267 I did something wrong....
oh, I see
2n^2+4-2n^2=268 2n^2-2n^2=264
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0=264?!?!
i made a mistake
\[(2n+2)^2-(2n)^2=8n+4\]
i forgot to square the 2!!
where didya get the "8n"?
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\[8n+4=268\] \[n=33\] \[2\times 33+2\] is your answer
algebra
\[(2n+2)(2n+2) -4n^2\] etc
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