Quantcast

A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

Jonask

  • 2 years ago

For how many positive integer values of c does the equation below have an integer solution?

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

    \[\huge \color{brown}{ 2x^2+689x+c}\]

  2. Jonask
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @TuringTest @phi @hartnn @sauravshakya @shubhamsrg

  3. Jonask
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[(2x+a)(x+b)=4x^2+2x(a+b)+ab\] \[2(a+b)=689,ab=c\]

  4. Jonask
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i dont kow wat to do here

  5. TuringTest
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @JamesJ D:

  6. Jonask
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    first 689 can never be written as even number

  7. TuringTest
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I am pretty bad at things like this for some reason, sorry.

  8. Jonask
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    thanks Turing

  9. Jonask
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    http://www.wolframalpha.com/input/?i=689

  10. Jonask
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    wolfram gives interger solutions as \[c=689n-2n^2, x=-n\]

  11. TuringTest
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I am assuming the equation is\[\large2x^2+689x+c=0\]right?

  12. Jonask
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    http://www.wolframalpha.com/input/?i=2x%5E2%2B689x%2Bc%3D0

  13. shubhamsrg
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    2x^2 + 689x + c = 0 c = -689x - 2x^2 x= n will always given a -ve c, hence x should be -n , for some integer n, => c = 689n - 2n^2 this has to be >0 => 689n -2n^2 >0 => n(689 - 2n) >0\ We already know n>0 hence we only have to deal with 689 > 2n or n < 344.5 Hence there can be 344 required value's of c. I have done this on brilliant.org if I am not wrong.

  14. Jonask
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yes its from them,i have many other problems from them,if you dont mind you can check the question i asked bfore this one...thanks makes sense

  15. Jonask
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    is this the only c

  16. shubhamsrg
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    What do you mean "the only c" ?

  17. Jonask
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    the question says For how many positive integer values of c so are there 344 interger c's

  18. shubhamsrg
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    yes.

  19. Jonask
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    okay thanks

  20. 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...

23

  • 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.