What is the minimum value of 'b' in the given equations?

[utkarshakash]

Homework Statement

Let a,b,c be in G.P. and a-b,c-a,b-c in H.P. If both roots of $(a+c)x^2 + bx + 4b^2=0$ 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
$\frac{2}{c-a}=\frac{1}{a-b}+\frac{1}{b-c} \\<br /> r^2+4r+1=0$

Since both roots are +ve sum and product of roots should also be +ve.
$\frac{-b}{a+c}>0 \\<br /> \frac{4b^2}{a+c}>0 \\ \\<br /> \frac{r}{1+r^2}<0\\<br /> \frac{4ar}{1+r^2}>0$

 

[haruspex]

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.

 

[utkarshakash]

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 $\dfrac{4ar^2}{1+r^2}$. Now if I change a to b/r it changes into$\dfrac{4br}{1+r^2}>0$. 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.

 

[haruspex]

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.

 

[utkarshakash]

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

 

[haruspex]

b(1+r)<1/16

I get something a little different. Pls post your working.

 

[utkarshakash]

I get something a little different. Pls post your working.

Sorry. I made a mistake. It should be $\dfrac{16br^2-r+16b}{r} \leq 0$. 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) $16br^2-r+16b \geq 0$. This means that discriminant of the equation in LHS is <=0. $b^2 \geq \frac{1}{256*4}$. From here minimum value of b is 1/32. Thus |[1/32]|=0. But this is not the answer.

 

[haruspex]

$16br^2-r+16b \geq 0$.

You already determined that r + 1/r = -4, so use that to eliminate r.

 

[utkarshakash]

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?

 

[haruspex]

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.