A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

ParthKohli

  • one year ago

I guess you guys will enjoy this problem (taken from Brilliant.org) -

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

    spit that fire

  2. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Find the sum of all integral values of \(\alpha \in [-20,20]\) such that\[2\log(x+3) = \log(\alpha x)\]has exactly one real solution.

  3. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    So first things first:\[x> -3\tag1\]\[\alpha x > 0 \tag 2 \]If we accept these two conditions, we can now recast the problem into the form.\[\log\left((x + 3)^2\right) = \log(\alpha x)\]\[\Rightarrow (x+3)^2 = \alpha x\]\[\Rightarrow x^2 + (6 - \alpha)x + 9 = 0\]Pay attention to the fact that the quadratic can have one real solution, but that is not the only possibility. The quadratic may as well have two real solutions, yet one of them making the logarithm undefined. So let's first see when the quadratic itself has one real solution. I'll save us the work and tell you guys that it occurs at \(\alpha = 0, 12\) but \(\alpha = 0\) is eliminated because of the second condition of course. Now comes the hard part in my opinion.

  4. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    |dw:1434998964257:dw|

  5. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    So let's try to check where \(x^2 + (6 - \alpha) x + 9\) has one real root lying in the interval \((\infty, -3]\). To have two real roots, \((6 - \alpha)^2 > 36\) o \(\alpha(\alpha - 12) > 0\) meaning \(\alpha \in (-\infty, 0) \cup (12, \infty)\). To have one root in the interval \((-\infty, -3)\) ...

  6. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    we know that \(f(-3) < 0\). Thus \(9 + 9 -3(6 - \alpha) < 0 \Rightarrow \alpha < 0\).

  7. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Ooo, interesting.

  8. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    That looks neat! for an alternative, im thinking of using calculus and mess with slopes the system has an unique solution when the slope of line is not so steep enough to cut the parabola at two different places and not too low to miss the parabola completely need to work integral values separately

  9. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    But if \(\alpha < 0\), we also know that the other root must be in the interval \((-3,0)\) since \(\alpha x>0\).

  10. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Do we really need to solve for the above condition? We already know that the roots of the equation \(x^2 + (6 - \alpha) x + 9\) have the same sign so wouldn't the second condition be covered?

  11. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    I'm not as experienced as you when it comes to calculus - but go right ahead.

  12. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    OK, \(\alpha = -20, -19, \cdots, -1, 12\) so far.

  13. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    @ikram002p Hey.

  14. ikram002p
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    hey pk :)

  15. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Where did Ganeshie leave?

  16. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    I'm missing steef. :(

  17. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    I think that's it - we can't really make the \(\log(\alpha x)\) undefined for one value and defined for the other since both roots have the same sign, right? So they're in this together.

  18. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Amazing, I got it right!

  19. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    That isn't as complicated as I thought it is going to be!

  20. ikram002p
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    :)

  21. 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.