SOLUTION: find two consecutive integers such that the square of the sum of the two integers is 8 more than the first integer.
Algebra
->
Problems-with-consecutive-odd-even-integers
-> SOLUTION: find two consecutive integers such that the square of the sum of the two integers is 8 more than the first integer.
Log On
Word Problems: Problems with consecutive odd even integers
Word
Solvers
Solvers
Lessons
Lessons
Answers archive
Answers
Click here to see ALL problems on Problems-with-consecutive-odd-even-integers
Question 954855
:
find two consecutive integers such that the square of the sum of the two integers is 8 more than the first integer.
Answer by
CubeyThePenguin(3113)
(
Show Source
):
You can
put this solution on YOUR website!
consecutive integers: x, (x+1)
(x + (x+1))^2 = x + 8
(2x+1)^2 = x + 8
4x^2 + 4x + 1 = x + 8
4x^2 + 3x - 7 = 0
(x - 1)(4x + 7) = 0
The integers are 1 and 2.
