SOLUTION: I am supposed to use applications of quadratic equations to solve this problem, I have worked on it about five hours now, and cannot figure it out. So I figure I am not translating

Question 148699This question is from textbook Beginning and Intermediate Algebra
: I am supposed to use applications of quadratic equations to solve this problem, I have worked on it about five hours now, and cannot figure it out. So I figure I am not translating the English correctly, could you please show me how to translate this.
The sum of the squares of two consecutive integers is nine less than ten times the sum of the integars. Find all such integers.
What I did says x^2(x+1)^2=10x+9 or I also used 10(x+9)
Please help me, this is making me nuts! This question is from textbook Beginning and Intermediate Algebra

Answer by ptaylor(2198) (Show Source): You can put this solution on YOUR website!
Let x=one integer; x^2 is the square of the integer
Then x+1=the other integer; (x+1)^2 is the square of the other integer
x+(x+1)=2x+1 is the sum of the integers 10(2x+1)=10 times the sum of the integers;10(2x+1)-9=9 less than 10 times the sum of the integer. So our equation to solve is:
x^2+(x+1)^2=10(2x+1)-9 simplify
x^2+x^2+2x+1=20x+10-9
2x^2+2x+1=20x+1 subtract 20x and also 1 from each side
2x^2+2x+1-1-20x=20x-20x+1-1 collect like terms
2x^2-18x=0 divide each term by 2
x^2-9x=0
x(x-9)=0
x=0; x+1=1
and
x=9; x+1=10
CK
0^2+1^2=10(0+1)-9
1=1
and
9^2+10^2=10(19)-9
81+100=190-9
181=181
Hope this helps---ptaylor

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