Question 948889
The product of two consecutive integers is 11  less than the square of the lesser integer. Find the greater of the two integers
x=lesser integer; x+1=greater integer
{{{(x)(x+1)=x^2-11}}}
{{{x^2+x=x^2-11}}} Subtract {{{x^2}}} from each side
x=-11 The lesser integer is -11
x+1=-11+1=-10 ANSWER: The greater integer is -10
CHECK:
{{{(-11)(-10)=(-11)^2-11}}}
{{{110=121-11}}}
{{{110=110}}}