document.write( "Question 715444: The product of two consecutive integers is less than the square of the smaller integer. Find the larger of the two integers.
\n" ); document.write( "The larger of the two integers is \n" ); document.write( "
Algebra.Com's Answer #439440 by fcabanski(1391)\"\" \"About 
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Call the two consecutive integers x and x+1. Their product x*(x+1) or \"x%5E2+%2B+x\" is less than the square of the smaller. The smaller is x, its square is \"x%5E2\" and \"x%5E2+%2B+x+%3C+x%5E2\"


\n" ); document.write( "Subtract \"x%5E2\" from both sides. x<0. Then the larger of the two integers x+1 < 1.


\n" ); document.write( "The problem has many solutions. For example if x=-1 then x+1 = 0 and the product 0*1 = 0 is less than the smaller squared = -1 squared = 1. 0 < 1.


\n" ); document.write( "If x=-10, then x+1 = -9. Product: 90. Smaller squared = \"-10%5E2\" = 100.


\n" ); document.write( "Let's try an x outside the solution, and as a result an x+1 outside the solution.


\n" ); document.write( "If x=0 then x+1 = 1. 0*1=0 and 0 squared = 0. The product is not less than the smaller number squared. That's what we expect since x+1 < 1 is the solution.
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