A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anonymous

  • one year ago

find all value of m such that sqrt(x) = m(x-2) has a solution greater than 4

  • This Question is Closed
  1. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    m>4 right

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

    Square both sides of the equation I guess

  3. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    are u looking for int solutions?

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

    @dan815 no, real numbers solution

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

    @Empty I guess you could do so. I've attempted and with a help with wolfram alpha, it got 0 < m < 1. But the arithmetic is just way tedious.

  6. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    ok well look at the behavior of x/(x-2)^2, find its critical, and inflection points also intersect it with y=4^2 LINE

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

    just wondering if there is any other easier method

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

    Solve for x, then set that value to be greater than 4. Hmmm Ohhh so you're looking for a trick, like a nice way to do it ok that makes this much more fun hahah ok hmmm give me a minute.

  9. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    oh so x>4 not m

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

    Just plug in x=4 and solve for m=1. Then you know m has to be greater than that, cause the square root is always positive, done.

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

    @dan815 yeah, it's about m that makes x > 4 as solution

  12. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    you can also think about it like a sqrt function intersecting with a line displaced 2 units right, and m is the slope

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

    @Empty huhm... so sqrt(4) = m(4-2), which gives m = 1. And then?

  14. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    |dw:1436389702432:dw|

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

    You want all solutions larger than that, so m>1. That's all your solutions. Here look: https://www.desmos.com/calculator/ryk0xxgqmf

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

    The case you solved for is the lowest possible case for the value of m. Well technically it's not because it's really m>1 not \(m \ge 1\)

  17. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    ya thatsa the simplest way

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

    @dan815 I thought of it that way too.

  19. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    just set x=4 and see what m has to be

  20. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    m has to be less than that value

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

    @Empty well, set = m = 1.1, then x is about 3.764 which is not greater 4

  22. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    he meant less

  23. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    m >0 and less than 1

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

    Yeah, as I said earlier, 0 < m < 1 but this is with the help of wolfram alpha.

  25. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    as a slope below 0 will make it intersect behind

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

    Right right sorry I am backwards in what I said

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

    XD haha sorry to be confusing

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

    I was imagining as m gets smaller the value of the point of their intersection increases.

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

    yeah, that helps confirm our intuition but as far as the algebra go, is there a nicer way to get 0 < m < 1?

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

    The tedious algebra is solving this, which wolfram alpha did https://www.wolframalpha.com/input/?i=%28%284m%5E2%2B1%29+%2B+sqrt%5B+%284m%5E2%2B1%29%5E2+-+4%28m%5E2%29%284m%5E2%29%5D+%29+%2F+%282m%5E2%29+%3E+4

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

    I'm about to go soon. So just work on it if you're interested.

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

    find all value of m such that sqrt(x) = m(x-2) has a solution greater than 4 $$\sqrt{x}=m(x-2)\\x=m^2(x-2)^2\\x=m^2(x^2-2x-4)\\m^2x^2-(2m^2+1)x-4=0\\x=\frac{2m^2+1\pm\sqrt{(2m^2+1)^2+16m^2}}{2m^2}$$to prove *a* solution greater than \(4\) all we need is:$$\frac{2m^2+1+\sqrt{(2m^2+1)^2+16m^2}}{2m^2}\ge4\\2m^2+1+\sqrt{(2m^2+1)^2+16m^2}\ge8m^2\\\sqrt{(2m^2+1)^2+16m^2}\ge6m^2-1\\(2m^2+1)^2 +16m^2\ge(6m^2-1)^2\\4m^4+4m^2+1+16m^2\ge 36m^4-12m^2+1\\32m^2\ge32m^4\\m^2(m^2-1)\le 0$$so our only solutions are where \(m^2=0\) or \(m^2\le 1\), which corresponds to \(|m|\le 1\)

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

    now, of course, some of these solutions are extraneous, since \(\sqrt{x}=m(x-2)\ge 0\) so \(m<0\) requires that \(x-2<0\implies x<2\) however we want solutions \(x\ge4\) we actually must have \( m>0\) so \(0\le m\le 1\)

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

    now, if you wanted to guarantee \(x>4\) strictly rather than \(x\ge 4\) then the solution is really \([0,1)\)

  35. 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.