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Minimum value of b

  • #1
utkarshakash
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Homework Statement


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]|

Homework Equations



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]
 

Answers and Replies

  • #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.
 
  • #3
utkarshakash
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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.
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.
 
  • #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.
 
  • #5
utkarshakash
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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.
b(1+r)<1/16
 
  • #6
haruspex
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b(1+r)<1/16
I get something a little different. Pls post your working.
 
  • #7
utkarshakash
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I get something a little different. Pls post your working.
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:
 
  • #8
haruspex
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[itex]16br^2-r+16b \geq 0[/itex].
You already determined that r + 1/r = -4, so use that to eliminate r.
 
  • #9
utkarshakash
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You already determined that r + 1/r = -4, so use that to eliminate r.
OK. This gives me the correct answer. But I want to know what was wrong in my solution?
 
  • #10
haruspex
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OK. This gives me the correct answer. But I want to know what was wrong in my solution?
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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