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Minimum value of b

  1. Oct 22, 2013 #1

    utkarshakash

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    1. The problem statement, all variables and given/known data
    Let a,b,c be in G.P. and a-b,c-a,b-c in H.P. If both roots of [itex](a+c)x^2 + bx + 4b^2=0[/itex] are positive and minimum value of 'b' be k then value of |[k]|

    2. Relevant equations

    3. The attempt at a solution
    Let a,b,c be denoted by a,ar,ar^2. now
    [itex]\frac{2}{c-a}=\frac{1}{a-b}+\frac{1}{b-c} \\
    r^2+4r+1=0[/itex]

    Since both roots are +ve sum and product of roots should also be +ve.
    [itex]\frac{-b}{a+c}>0 \\
    \frac{4b^2}{a+c}>0 \\ \\
    \frac{r}{1+r^2}<0\\
    \frac{4ar}{1+r^2}>0[/itex]
     
  2. jcsd
  3. Oct 22, 2013 #2

    haruspex

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    I think the left hand term in the last line of your post is wrong.
    Can you deduce the sign of a?
    It might help to express everything in terms of b and r rather than a and r, since the question concerns b, not a.
     
  4. Oct 23, 2013 #3

    utkarshakash

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    It should be [itex]\dfrac{4ar^2}{1+r^2}[/itex]. Now if I change a to b/r it changes into[itex]\dfrac{4br}{1+r^2}>0[/itex]. Now denominator will always be +ve so I can simply write br>0. Also from the first inequality I can say that r<0. To satisfy both inequalities b must be less than zero. Now b=ar<0. Thus a must be +ve.
     
  5. Oct 23, 2013 #4

    haruspex

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    The info in the OP implies both roots are real, since a complex root cannot be said to be positive or negative. You haven't used that.
     
  6. Oct 23, 2013 #5

    utkarshakash

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    b(1+r)<1/16
     
  7. Oct 23, 2013 #6

    haruspex

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    I get something a little different. Pls post your working.
     
  8. Oct 23, 2013 #7

    utkarshakash

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    Sorry. I made a mistake. It should be [itex]\dfrac{16br^2-r+16b}{r} \leq 0[/itex]. I have used the fact that discriminant of the original equation is greater than 0 and substituting a as b/r. As r<0 (deduced earlier) [itex]16br^2-r+16b \geq 0[/itex]. This means that discriminant of the equation in LHS is <=0. [itex]b^2 \geq \frac{1}{256*4}[/itex]. From here minimum value of b is 1/32. Thus |[1/32]|=0. But this is not the answer. :frown:
     
  9. Oct 24, 2013 #8

    haruspex

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    You already determined that r + 1/r = -4, so use that to eliminate r.
     
  10. Oct 24, 2013 #9

    utkarshakash

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    OK. This gives me the correct answer. But I want to know what was wrong in my solution?
     
  11. Oct 24, 2013 #10

    haruspex

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    Because you did not use all the available information you did not get a strong enough bound. Note that even now you have not proved your answer to be correct. To do that you would need to demonstrate that every b > k occurs in a solution of the given equations.
     
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