Question

The product of two consecutive positive integers is greater than their sum by 209. Find these numbers.

Answer 1

To solve this problem, set up the equation x(x+1) = x + (x+1) + 209. Simplify and factor the quadratic equation to find the possible values for x. Choose the positive value for x as the solution.

Step-by-step explanation:

To solve this problem, let's assume the two consecutive positive integers are x and x+1. The product of these two numbers is x(x+1) and their sum is x + (x+1). According to the given information, the product is greater than the sum by 209, so we can set up the equation x(x+1) = x + (x+1) + 209.

Expanding and simplifying the equation, we get x² + x = 2x + 210.

Moving the terms to one side, we have x² - x - 210 = 0. Factoring the quadratic equation, we get (x-15)(x+14) = 0.

This gives us two possible values for x: x = 15 or x = -14. Since we're looking for positive integers, the solution is x = 15. Therefore, the consecutive positive integers are 15 and 16.

Answer 2

15, 16

Step-by-step explanation:

so, x(x+1)=x+x+1+209.solve this and you get x^2+x=2x=210, x^2-x-210=0, and when you solve this quadratic function, you get x=15, and the other with 16. to check see if 240 is 209 greater than 31, which is so there you have it. Hope you have a happy thanksgiving!