Question

Solve each of the following quadratic equations:

$\frac{x}{x-1}+\frac{x-1}{x}=4\frac{1}{4},x\ne 0,1$ [CBSE 2011]

Answer 1

$$\begin{gathered} \frac{x}{x-1}+\frac{x-1}{x}=4\frac{1}{4},x\ne 0,1 \\ \Rightarrow \frac{x^2+{\left(x-1\right)}^2}{x\left(x-1\right)}=\frac{17}{4} \\ \Rightarrow \frac{x^2+x^2-2x+1}{x^2-x}=\frac{17}{4} \\ \Rightarrow \frac{2x^2-2x+1}{x^2-x}=\frac{17}{4} \end{gathered}$$
$$\begin{gathered} \Rightarrow 8x^2-8x+4=17x^2-17x \\ \Rightarrow 9x^2-9x-4=0 \\ \Rightarrow 9x^2-12x+3x-4=0 \\ \Rightarrow 3x\left(3x-4\right)+1\left(3x-4\right)=0 \end{gathered}$$
$$\begin{gathered} \Rightarrow \left(3x-4\right)\left(3x+1\right)=0 \\ \Rightarrow 3x-4=0\mathrm{or}3x+1=0 \\ \Rightarrow x=\frac{4}{3}\mathrm{or}x=-\frac{1}{3} \end{gathered}$$
Hence, $\frac{4}{3}$ and $-\frac{1}{3}$ are the roots of the given equation.