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Find the non-zero value of c for which there is exactly one positive value of b for which there is one solution to the equation \(x^2 + \left(b + \frac 1b\right)x + c = 0.\)

 Apr 20, 2019
edited by Lightning  Apr 20, 2019
edited by Lightning  Apr 20, 2019
 #2
avatar+7763 
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\(\text{When }ax^2 + bx + c = 0\text{ has only 1 solution, }b^2 = 4ac\)

So 

\(\left(b+\dfrac 1 b\right)^2 = 4c\\ b^2 + \dfrac 1 {b^2} - 4c + 2 = 0\\ \left(b^2\right)^2 - (4c - 2)b^2 +1 = 0\\ \)

And there is only 1 positive value for b, so there are only 1 possible value for b.

Use that again and you get: 

\(\left(-(4c-2)\right)^2 = 4\\ 2c - 1 = 1\\ c = 1\)

 

I am not sure about what the question is talking about, but I guess you meant that. 

 Apr 21, 2019

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