Find 2 consecutive odd number whose product is 224

Answer 1

Let's assume the two consecutive odd numbers are n and n + 2.

We are given that the product of these two numbers is 224. Therefore, we can set up the following equation:

n(n + 2) = 224

Expanding the equation, we get:

n^2 + 2n = 224

Rearranging the equation and simplifying, we have:

n^2 + 2n - 224 = 0

To solve this quadratic equation, we can either factor it or use the quadratic formula. In this case, factoring is not feasible, so we will use the quadratic formula:

n = (-b ± √(b^2 - 4ac)) / (2a)

For this equation, a = 1, b = 2, and c = -224. Substituting these values into the quadratic formula, we can solve for n.

n = (-2 ± √(2^2 - 4(1)(-224))) / (2(1))

n = (-2 ± √(4 + 896)) / 2

n = (-2 ± √900) / 2

n = (-2 ± 30) / 2

Now we have two possible values for n:

n = (-2 + 30) / 2 = 28 / 2 = 14

n = (-2 - 30) / 2 = -32 / 2 = -16

Therefore, the two consecutive odd numbers whose product is 224 are 14 and 16.