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)![]() ![]() ![]() You can put this solution on YOUR website! 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( " |