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Got stuck due to the inequality not being satisfied

  1. Nov 12, 2016 #1
    1. The problem statement, all variables and given/known data
    Let ##a,b,c## be positive integers and consider all the quadratic equations of the form ##ax^2-bx+c=0## which have two distinct real roots in ##(0,1)##. Find the least positive integers ##a## and ##b## for which such a quadratic equation exist.

    2. Relevant equations


    3. The attempt at a solution
    Let ##f(x)=ax^2-bx+c##. Now as the roots of ##f(x)=0## lie in ##(0,1)## hence ##f(0)\cdot f(1)\gt0##.

    $$D\gt0\implies b^2-4ac\gt0$$
    $$f(0)=c \\ f(1)=a-b+c$$

    As, ##a,b,c\in\mathbb{Z}^+##, so
    $$f(0)\ge1 \implies c\ge1\tag{1}$$
    $$f(1)\ge1\implies a-b+c\ge1\tag{2}$$
    $$f(0)\cdot f(1)\ge1\implies c(a-b+c)\ge1\implies c^2+(a-b)c-1\ge0\tag{3}$$

    Now, for the quadratic inequality in ##c## that we obtained in ##(3)##, for it to hold we get

    $$D\le0\implies (a-b)^2+4\le0$$
    The above inequality doesn't hold ##\forall a,b \in \mathbb{Z}^+##.

    After arriving at this conclusion I am kind of stuck on what to conclude from the above result or rather am I going in the right direction.

    A little push in the right direction in the form of a hint or rather what should be the line of thought for attempting the question. Please don't write the solution, just the line of thought to arrive at the solution would be enough.
     
  2. jcsd
  3. Nov 12, 2016 #2
    @haruspex. Can you guide me in the right direction.
     
  4. Nov 12, 2016 #3

    fresh_42

    Staff: Mentor

    Why? And why don't you analyse the roots by the formula for them?
     
  5. Nov 12, 2016 #4
    I don't think it gives a stricter bound.
     
  6. Nov 12, 2016 #5

    fresh_42

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    I haven't done the exercise, but there is at least another condition, because you not only have ##D > 0## but also ##0 < b \pm \sqrt{D} < 2a\,##.
     
  7. Nov 12, 2016 #6
    Yeah and on solving bot the inequalities the cases that you get are
    $$4a(a-b+c)\gt0 \\ 4ac\gt0$$ and we already have stricter bounds than these with $$f(1)\ge1\implies a-b+c\ge1 \\ f(0)\ge1\implies c\ge1$$ which I have already shown in my original post.
     
    Last edited: Nov 12, 2016
  8. Nov 12, 2016 #7

    fresh_42

    Staff: Mentor

    How did you get
    from ##(3)##? Isn't e.g. ##(15,15,1)## one possible solution?
     
  9. Nov 12, 2016 #8
    Yeah that's right, and I got the solution. will be done writing it in my notes, then will post it. The inequalities that need to be solved are:
    $$|a|\gt4$$
    $$c\lt a$$
    $$a+c\lt b$$
    $$b^2\gt4ac$$

    From these we get the least value of ##a## and ##b## ##(a,b)=(5,5)##

    Will be posting a thorough solution in about half an hour.
     
  10. Nov 12, 2016 #9
    After a thinking a lot on the question I was able to come up with a good enough solution.

    Let ##f(x)=ax^2-bx+c##. Let the roots of the equation ##f(x)=0## be ##\alpha## and ##\beta##. Then, ##\alpha,\beta\in(0,1)##.

    As the roots of the equation ##f(x)=0## are real and distinct, then ##b^2\gt4ac##. Also, due to the roots lying in ##(0,1)##, hence ##f(0)\gt0##, ##f(1)\gt0## and ##f(0)f(1)\gt0##

    Now, $$f(0)=c\ge1\tag{1}$$
    $$f(1)=a-b+c\ge1\tag{2}$$
    $$\therefore \qquad f(0)\cdot f(1)\ge1\qquad\qquad\tag{3}$$
    $$b^2\gt4ac\tag{4}$$

    If ##\alpha## and ##\beta## are the zeroes of ##f(x)##, then ##f(x)=a(x-\alpha)(x-\beta)##
    $$\therefore f(0)\cdot f(1)=a\alpha(1-\alpha)a\beta(1-\beta)\ge 1$$

    Using A.M-G.M inequality we get,
    $$\dfrac{b+(1-b)}{2}\ge\sqrt{b(1-b)}\implies b(1-b)\le\dfrac{1}{4}$$
    equality occurs when ##b=\dfrac{1}{2}##
    $$\therefore \qquad\qquad\alpha(1-\alpha)\ge\dfrac{1}{4}\qquad\qquad \beta(1-\beta)\ge\dfrac{1}{4}$$

    In both the cases equality occurs when ##\alpha=\beta=\dfrac{1}{2}##, but as ##\alpha\neq\beta##

    So, $$\alpha(1-\alpha)\beta(1-\beta)\gt\dfrac{1}{16}\implies a\alpha(1-\alpha)a\beta(1-\beta)\gt\dfrac{a^2}{16}\\
    \implies \dfrac{a^2}{16}\lt f(0)\cdot f(1)\implies |a|\lt4$$

    $$\therefore a_{min}=5$$

    Also, $$\because \qquad\qquad\alpha\cdot\beta\lt1\implies c\lt a_{min}\qquad\qquad\tag{5}$$

    $$(2)\implies a+c\ge1+b\implies a_{min}+c-1\ge b\implies 4+c\ge b \\
    \implies c^2+8c+16\ge b^2\implies c^2+8c+16 \gt 4a_{min}c\\
    \implies c^2-12c+16\gt0\implies c\in(0,6-2\sqrt5)\cup(6+2\sqrt5,\infty)$$

    But, as ##c\lt a_{min}##, hence ##c=1##

    Now, from $$b^2\gt 4a_{min}c\implies b^2\gt20$$
    Hence, ##b_{min}=5##

    So, the value of least positive integers ##a##, ##b## and ##c## for which ##f(x)## has roots b/w ##0## and ##1## is ##(a,b,c)=(5,5,1)##
     
  11. Nov 12, 2016 #10

    haruspex

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    I don't follow that step. You had f(0)f(1)>=1, not f(0)f(1)<=1.
    If you meant |a|>4, that does not follow either.
     
  12. Nov 13, 2016 #11
    ##ax^2 -bx + c = f(x)##
    ##f(0) = c \ge 1##
    ##f(1) = a-b+c \ge 1##
    ##D \gt 0 \implies b^2 -4ac \gt 0 \implies b^2 \gt 4ac##


    Putting it together
    ##c \ge 1##
    ##a-b+c \ge 1 \implies a+c \ge b+1##
    ##b^2 \gt 4ac##


    Taking ##c = 1##

    We get,
    ##a \ge b##
    ##b^2 \gt 4a##

    For lowest value we fix ##a = b##
    ##b^2 \gt 4b##
    From which we get, ##b \gt 4##
    For lowest value, ##a = b = 5##
    So solution is ##(a,b,c)=(5,5,1)##

    @haruspex
    @fresh_42

    Is this correct ?
     
    Last edited: Nov 13, 2016
  13. Nov 13, 2016 #12

    fresh_42

    Staff: Mentor

    Except I don't know what A.M-G.M inequality is, where you lost me in post #9 so I couldn't see why ##b(1-b)\leq\frac{1}{4}##,
    I have the following questions: The inequalities ##b^2 > 4ac\, , \,c \geq 1\, , \,a+c \geq b+1## you have are o.k. Now you set ##c=1## and next ##a=b## which gave you ##(a,b,c)=(5,5,1)## which is a solution and minimal under these conditions. Right.

    But what happens in the cases
    1. ##c=1 \,\wedge\, a>b##?
    2. ##c > 1##?
    Since it is not defined what the least value of a triple ##(a,b,c)## means (sum, average, least maximal value, least minimal value or whatever), one has to somehow show, that the cases above are either impossible or lead to higher numbers of all three values ##a,b,c##. The first case is easy (by looking at ##ab > b^2 > 4a##), but why can't e.g. ##(a,b,c)=(3,4,2)## or something like this be a solution?
     
  14. Nov 13, 2016 #13

    haruspex

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    The OP actually says the least a and b; c is not required to be minimised. (But as you say, that is still not well defined.)
    Consequently...
    ... is not justified.
     
  15. Nov 13, 2016 #14
    I don't have a proof but condition for a, b to minimum for a given c is ##a=b=1+4c## where ##c > 0##.
     
  16. Nov 13, 2016 #15

    haruspex

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    If you are relying on that then it needs to be demonstrated within your proof prior to the line "Taking c=1".
     
  17. Nov 13, 2016 #16
    Yes I do rely on that but I don't have a proof.
    I observed it while playing with a graphing calculator. Can you prove/disprove it ?
     
  18. Nov 14, 2016 #17
    Yeah, that was a mistake on my part while writing the solution and the chain errors occured because I was not following the rough work I had done but what I had typed on the screen, so it should be
    $$\alpha(1-\alpha)\beta(1-\beta)\lt\dfrac{1}{16}\implies a\alpha(1-\alpha)a\beta(1-\beta)\lt\dfrac{a^2}{16}\\
    \implies \dfrac{a^2}{16}\gt f(0)\cdot f(1)\implies |a|\gt4$$
     
  19. Nov 14, 2016 #18
    I just wrote the triplet ##(a,b,c)## to make it easy to write the values that correspond to each ##a,b## and ##c## it just resembles and ordered pair.

    Although the question doesn't ask for the least value of ##c## but if it is not minimised then we will not get the least value of all the possible values of ##a## and ##b## for a different ##c##
     
    Last edited: Nov 14, 2016
  20. Nov 14, 2016 #19

    Mark44

    Staff: Mentor

    Don't write stuff like $(a, b, c)$ -- it doesn't do anything but clutter up what you wrote. We have a LaTeX tutorial here: https://www.physicsforums.com/help/latexhelp/
     
  21. Nov 14, 2016 #20
    Oh, sorry I was just writing some stuff in TeX editor hence forgot that we are supposed to use "##" here instead of "$" for inline.
     
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