Question

Find two positive consecutive odd integers such that the square of the large integer is one less than twice the square of the smaller integer.

Answer 1

Step-by-step explanation:

x, x+2 are the two positive consecutive of integers (x + 2, because the minimum distance between 2 odd numbers is always 2, as there can never be 2 odd numbers directly following each other; and neither can even numbers, by the way).

(x+2)² = 2×x² - 1

x² + 4x + 4 = 2x² - 1

0 = x² - 4x - 5

we can solve this by factoring.

that means we need to find a and b so that

(x - a)(x - b) = x² - 4x - 5 = 0

and x = a and x = b are the 2 solutions of the quadratic equation.

x² - ax - bx + ab = x² - 4x - 5 = 0

x² - (a+b)x + ab = x² - 4x - 5 = 0

a+b = 4

ab = -5

that means that either a or b has to be 5 and the other -1.

so, the result is

(x - 5)(x + 1) = 0

x = 5

or

x = -1

are the possible solutions.

since we are only looking for positive integers, x = -1 is not a valid solution.

and only x = 5 remains as valid solution.

therefore, the 2 positive consecutive odd integers are

5, 7