A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing


  • 5 years ago

Find four consecutive intergers such that the product of the 1st interger and the 3rd interger is 25 greater than the product of -13 and the fourth integer Answers: -14,-13,-12,-11

  • This Question is Closed
  1. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Let’s say that your sequence of four numbers starts at some number n. Then the four numbers are n, n+1, n+2, and n+3. You’re asked to find that number n so that n(n+2) = 25 + (-13)(n+3). One way to solve this is to notice that this is a quadratic equation in n. You’re guaranteed that an integer solution exists, so you’d expect to be able to factor the quadratic expression. If you try, you’ll simplify it down to n^2 + 15n + 14 = 0. And this is (n + 14)(n + 1) = 0. So n can be either -14 or -1. If n = -14, then you get the sequence -14, -13, -12, -11. If you let n = -1, then you get the sequence -1, 0, 1, 2. You can check that both sequences satisfy our condition on the numbers.

  2. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...


  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.