SOLUTION: the square of the positive integer is 98 less than twice the square of the next consecutive positive integer. what are the integers?

Algebra ->  Problems-with-consecutive-odd-even-integers -> SOLUTION: the square of the positive integer is 98 less than twice the square of the next consecutive positive integer. what are the integers?       Log On


   

Question 474856: the square of the positive integer is 98 less than twice the square of the next consecutive positive integer. what are the integers?

Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
the square of the positive integer is 98 less than twice the square of the next consecutive positive integer.
what are the integers?
:
Write an equation for what it says:
x^2 = 2(x+1)^2 - 98
x^2 = 2(x^2+2x+1) - 98
x^2 = 2x^2 + 4x + 2 - 98
0 = 2x^2 - x^2 + 4x - 96
A quadratic equation
x^2 + 4x - 96 = 0
Factors to
(x+12)(x-8) = 0
the positive solution
x = 8 and 9 are the integers
:
:
Check this
8^2 = 2(9^2) - 98