document.write( "Question 273359: There is three consecutive positive integers. if the square of the third integer is 51 more than the sum of the first two what are the integers? \n" ); document.write( "
Algebra.Com's Answer #199737 by checkley77(12844)\"\" \"About 
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LET X, X+1 & X+2 BE THE 3 INTEGERS.
\n" ); document.write( "(X+2)^2=X+X+1+51
\n" ); document.write( "X^2+4X+4=2X+52
\n" ); document.write( "X^2+4X-2X+4-52=0
\n" ); document.write( "X^2+2X-48=0
\n" ); document.write( "(X-6)(X+8)=0
\n" ); document.write( "X-6=0
\n" ); document.write( "X=6 ANS. FOR THE FIRST INTEGER.
\n" ); document.write( "6+1=7 ANS. FOR THE SECOND INTEGER.
\n" ); document.write( "6+2=8 FOR THE LARGEST INTEGER.
\n" ); document.write( "PROOF:
\n" ); document.write( "8^2=6+7+51
\n" ); document.write( "64=64
\n" ); document.write( "
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