Find two positive consecutive odd integers such that the square of the larger integer is one less than twice the square of the small integer.
Accepted Solution
A:
Let x---------------> first positive odd integer x+2------------> second positive consecutive odd integer
we know that (x+2)²=2x²-1---------> x²+4x+4=2x²-1-------> x²-4x-5 x²-4x-5=0 using a graph tool--------> to calculate the quadratic equation see the attached figure
the solution is x=5
the answer is the first positive odd integer x is 5 the second positive consecutive odd integer x+2 is 7