document.write( "Question 161445This question is from textbook Introductory Algebra
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document.write( ": Find two consectutive intergers such that the square of the sum of the two intergers is 11 more than the smaller interger.\r
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document.write( "I tried to set this up like : x(x+1)=(x+x+1)+11\r
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document.write( "and the answer is -2,-1.\r
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document.write( "but of course we have to show our work ..... \n" );
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Algebra.Com's Answer #118914 by checkley77(12844) ![]() You can put this solution on YOUR website! (X+X+1)^2==X+11 \n" ); document.write( "(2X+1)^2=X+11 \n" ); document.write( "4X^2+4X+1=X+11 \n" ); document.write( "4X^2+4X-X+1-11=0 \n" ); document.write( "4X^2+3X-10=0 \n" ); document.write( "(4X-5)(X+2)=0 \n" ); document.write( "4X-5=0 \n" ); document.write( "4X=5 \n" ); document.write( "X=5/4=1.25 ANSWER. \n" ); document.write( "X=2=0 \n" ); document.write( "X=-2 ANSWER. \n" ); document.write( " |