Question

Solve each of the following quadratic equations:

(i) $\frac{x}{x-1}+\frac{x-1}{x}=4\frac{1}{4},x\ne 0,1$
(ii) $\frac{x-1}{2x+1}+\frac{2x+1}{x-1}=2,x\ne -\frac{1}{2},1$ [CBSE 2017]

Answer 1

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

(ii)
$$\begin{gathered} \frac{x-1}{2x+1}+\frac{2x+1}{x-1}=2 \\ \Rightarrow \frac{(x-1{)}^2+(2x+1{)}^2}{(2x+1)(x-1)}=2 \\ \Rightarrow (x^2+1-2x)+(4x^2+1+4x)=2(2x+1)(x-1) \\ \Rightarrow 5x^2+2x+2=2(2x^2-x-1) \\ \Rightarrow 5x^2+2x+2=4x^2-2x-2 \\ \Rightarrow x^2+4x+4=0 \\ \Rightarrow x^2+2x+2x+4=0 \\ \Rightarrow x(x+2)+2(x+2)=0 \\ \Rightarrow (x+2)(x+2)=0 \\ \Rightarrow (x+2)=0\mathrm{or}(x+2)=0 \\ \Rightarrow x=-2\mathrm{or}x=-2 \\ \Rightarrow x=-2 \end{gathered}$$