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 \(x^2 + \left(b + \frac 1b\right)x + c = 0\)

Guest Jun 1, 2019

edited by
Guest
Jun 1, 2019

#1**+2 **

There is only one solution for x when the discrimant is 0 , that is, when...

\((b+\frac1b)^2-4(1)(c)\ =\ 0\\~\\ (b+\frac1b)^2-4c\ =\ 0\\~\\ (b+\frac1b)(b+\frac1b)-4c\ =\ 0\\~\\ b^2+2+\frac{1}{b^2}-4c\ =\ 0\\~\\ b^2+(2-4c)+\frac{1}{b^2}\ =\ 0\\~\\ b^4+(2-4c)b^2+1\ =\ 0\qquad\text{Let}\qquad u=b^2\\~\\ u^2+(2-4c)u+1\ =\ 0\)

There is only one solution for u, and thus only one *positive *value of b , when...

\((2-4c)^2-4\ =\ 0\\~\\ (2-4c)^2\ =\ 4\\~\\ 2-4c\ =\ 2\qquad\text{or}\qquad2-4c\ =\ -2\\~\\ c\ =\ 0\phantom{2-4}\qquad\text{or}\qquad c\ =\ 1\)

The non-zero value of c is 1 .

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To check this answer, let's find the values of b for which \(x^2+(b+\frac1b)x+1=0\) has only one solution.

\(x^2+(b+\frac1b)x+1=0\) has only one solution when....

\((b+\frac1b)^2-4\ =\ 0\\~\\ b^2-2+\frac{1}{b^2}\ =\ 0\\~\\ b^4-2b^2+1\ =\ 0\\~\\ (b^2-1)^2\ =\ 0\\~\\ b\ =\ \pm1\)

There is only one positive value of b for which there is one solution to the equation \(x^2+(b+\frac1b)x+1=0\)

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hectictar Jun 1, 2019