Given that the equation $\frac{x}{x + 1} + \frac{x + 1}{x} = \frac{4x + a}{x(x + 1)},a \in R$, has o

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Given that the equation $\frac{x}{x + 1} + \frac{x + 1}{x} = \frac{4x + a}{x(x + 1)},a \in R$, has only one real root, then

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$x^{2} + (x + 1)^{2} - 4x - a = 0$$$\begin{matrix} & \ \Rightarrow \ 2x^{2} - 2x + 1 - a = 0 \ & D \geq 0\ \Rightarrow \ 4 + 8(a - 1) > 0 \Rightarrow a - 1 \geq - \frac{1}{2}\ \Rightarrow a \geq \frac{1}{2} \end{matrix}$$Put $x = 0,a = 1$$$\begin{matrix} & x = - 1,a = 5 \ & D = 0,a = \frac{1}{2} \end{matrix}$$Hence, $a = 1,5,\frac{1}{2}$

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