SOLUTION: Find two consectutive intergers such that the square of the sum of the two intergers is 11 more than the smaller interger. I tried to set this up like : x(x+1)=(x+x+1)+11

Algebra ->  Problems-with-consecutive-odd-even-integers -> SOLUTION: Find two consectutive intergers such that the square of the sum of the two intergers is 11 more than the smaller interger. I tried to set this up like : x(x+1)=(x+x+1)+11       Log On


   

Question 161445This question is from textbook Introductory Algebra
: Find two consectutive intergers such that the square of the sum of the two intergers is 11 more than the smaller interger.


I tried to set this up like : x(x+1)=(x+x+1)+11
and the answer is -2,-1.
but of course we have to show our work ..... This question is from textbook Introductory Algebra

Found 2 solutions by checkley77, scott8148:
Answer by checkley77(12844) About Me  (Show Source):
You can put this solution on YOUR website!
(X+X+1)^2==X+11
(2X+1)^2=X+11
4X^2+4X+1=X+11
4X^2+4X-X+1-11=0
4X^2+3X-10=0
(4X-5)(X+2)=0
4X-5=0
4X=5
X=5/4=1.25 ANSWER.
X=2=0
X=-2 ANSWER.

Answer by scott8148(6628) About Me  (Show Source):
You can put this solution on YOUR website!
" the square of the sum of the two intergers is 11 more than the smaller interger"
__ (x+x+1)^2=x+11 __ (2x+1)^2=x+11 __ 4x^2+3x-10=0 __ factoring __ (4x-5)(x+2)=0

4x-5=0 __ x=5/4 __ not an integer

x+2=0 __ x=-2 __ x+1=-1