SOLUTION: The sum of the squares of two consecutive integers is 41. What is the larger integer?

Question 101498: The sum of the squares of two consecutive integers is 41. What is the larger integer?
Answer by edjones(8007) (Show Source): You can put this solution on YOUR website!
let the integers be x and x+1
x^2 + (x+1)^2=41
x^2+x^2+2x+1=41
2x^2+2x-40=0
2(x^2+x-20)=0
2(x+5)(x-4)=0 Factor
x=4 , x=-5
the problem has two answers.
-5, -4
and 5, 4
4 and -5 the smaller integers.
x+1= 5 and -4 are the larger integers. ANSWER
Ed
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