1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Inequality problem

  1. Jul 2, 2013 #1
    1. The problem statement, all variables and given/known data
    Find all numbers ##a## for each of which the least value of the quadratic trinomial ##4x^2-4ax+a^2-2a+2## on the interval ##0\leq x \leq 2## is equal to 3.


    2. Relevant equations



    3. The attempt at a solution
    I don't really know what should be the best way to start with this type of question. I started with finding out the roots of the given equation. The roots come out to be
    [tex]x_1=\frac{a+\sqrt{2(a-1)}}{2}, x_2=\frac{a-\sqrt{2(a-1)}}{2}[/tex]
    The roots exist when ##a>1## and it is easy to see that ##x_1>x_2## for ##a>1##. I don't know how to proceed ahead. 0 and 2 should not lie between the roots as we need the minimum value to be 3. Also, ##x_1## is always greater than zero for ##a>1## so 0 and 2 should lie before ##x_1## and ##x_2## on a number line. Hence, the least value will be at ##x=2## and this should be equal to 3.
    [tex]4(2)^2-4a(2)+a^2-2a+2=3[/tex]
    Solving this, I get ##a=5+\sqrt{10}, 5-\sqrt{10}##. Both the values are greater than hence both are admissible but the answer key does not show up the second value i.e ##5-\sqrt{10}##. :confused:

    Any help is appreciated. Thanks!
     
    Last edited: Jul 2, 2013
  2. jcsd
  3. Jul 2, 2013 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi Pranav-Arora! :smile:

    You seem to be misinterpreting the question.

    The question isn't about the roots, it's about the minimum. :confused:
     
  4. Jul 2, 2013 #3
    I understand that the question asks me about the minimum. :smile:

    0 and 2 should not lie between the roots. The given minimum is 3 which is a positive value, and between the roots, the value of quadratic is negative.

    But I am still not sure about how should I begin? What I posted were my thoughts about approaching the problem but I feel that is incorrect. Can I have a few hints? :)
     
  5. Jul 2, 2013 #4

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    i'd start by rewriting the original equation in the form (x - b)2 - c2 ≥ 0 :wink:
     
  6. Jul 2, 2013 #5
    Sorry tiny-tim, there was a typo in the first post, I have corrected it. The correct quadratic is ##4x^2-4ax+a^2-2a+2##, I accidentally wrote -2x instead of -2a. Should I still follow your hint? Sorry for the trouble. :redface:
     
  7. Jul 2, 2013 #6

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

  8. Jul 2, 2013 #7
    I came up with this: ##(2x-a)^2-2(a-1)## but this doesn't match up with what you asked me. And how did you get that greater than or equal to symbol? :confused:
     
  9. Jul 2, 2013 #8

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    and the minimum of that (for 0 ≤ x ≤ 2) is … ? :smile:
    well, we want a minimum, and the question is posed as an inequality problem …

    so i thought it was time to throw in an inequality! :biggrin:
     
  10. Jul 2, 2013 #9
    ##a^2-2(a-1)##

    :confused:
    Why greater than or equal to zero? Why not greater than or equal to 3? I guess I am missing something really basic. :uhh:
     
  11. Jul 2, 2013 #10

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    i preferred to get everything over onto the LHS (and only 0 on the RHS) :smile:
     
  12. Jul 2, 2013 #11

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    At some stage I think you're going to have to break it into cases. You're looking for the minimum only within the range [0, 2]. In the OP you considered the possibility that the minimum occurs at x=2. You found two values for a that give f(2)=3 , but you did not check whether either of these give f(x)>=3 for all x in [0, 2].
    You also seem to have ruled out other locations for the minimum erroneously. There are two cases to consider for alternative locations of the minimum.
     
  13. Jul 2, 2013 #12
    tiny-tim, I still do not understand your inequality method but meanwhile, I tried something else and reached the answer but not sure if my approach is correct.

    I had ##(2x-a)^2-2(a-1)##. The squared term is zero when x=a/2.

    Case I:
    ##0 \leq a/2 \leq 2 \Rightarrow 0 \leq a \leq 4##
    For this case, the minimum value is at a/2 i.e ##-2(a-1)##. This is equal to 3 when a=-1/2 but this is not possible as a ranges between 0 and 4.

    Case II:
    When ##a/2 > 2 \Rightarrow a > 4##, the minimum value is at 2.
    Solving the quadratic, I get ##a=5+\sqrt{10}, 5-\sqrt{10}##. Second value is not possible as a is greater than 4.

    Case III:
    When ##a/2 < 0 \Rightarrow a<0##, the minimum value is at 0. From this I get two values ##a=1-\sqrt{2}, 1+\sqrt{2}##. Only the first value is possible.

    Hence, ##a=1-\sqrt{2}, 5+\sqrt{10}##. This is the correct answer however I am unsure about my approach.

    I am still interested in understanding tiny-tim's method.
     
  14. Jul 2, 2013 #13

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    That looks good. I broke it into cases according to whether the min was at 0, 2, or in-between, but the two methods are pretty much equivalent. I doubt tiny-tim's method was much different either, but it would be interesting to see.
     
  15. Jul 4, 2013 #14
    Bump...
     
  16. Jul 5, 2013 #15

    verty

    User Avatar
    Homework Helper

    I believe Tiny-tim was using calculus to get the answer because he had you put it in the best form to apply calculus to solve it. I won't give that method though. As a precalculus question, using the axis of symmetry is the most intuitive way, I believe. And Tim's final formula is the turning point form, so he might well have been hinting that you think about the axis of symmetry or use calculus, which I think is absolutely correct.

    Okay?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Inequality problem
  1. Inequality problem (Replies: 3)

  2. Inequality problem (Replies: 11)

  3. Inequality problem (Replies: 2)

  4. Inequality problem (Replies: 1)

  5. Inequality problem (Replies: 11)

Loading...