Question

Solve each of the following quadratic equations:

(i) $\frac{1}{x-1}-\frac{1}{x+5}=\frac{6}{7},x\ne 1,-5$
(ii) $\frac{1}{2x-3}+\frac{1}{x-5}=1\frac{1}{9},x\ne \frac{3}{2},5$ [CBSE 2017]

Answer 1

(i)
$$\begin{gathered} \frac{1}{x-1}-\frac{1}{x+5}=\frac{6}{7},x\ne 1,-5 \\ \Rightarrow \frac{x+5-x+1}{\left(x-1\right)\left(x+5\right)}=\frac{6}{7} \\ \Rightarrow \frac{6}{x^2+4x-5}=\frac{6}{7} \\ \Rightarrow x^2+4x-5=7 \end{gathered}$$
$$\begin{gathered} \Rightarrow x^2+4x-12=0 \\ \Rightarrow x^2+6x-2x-12=0 \\ \Rightarrow x\left(x+6\right)-2\left(x+6\right)=0 \\ \Rightarrow \left(x+6\right)\left(x-2\right)=0 \end{gathered}$$
$$\begin{gathered} \Rightarrow x+6=0\mathrm{or}x-2=0 \\ \Rightarrow x=-6\mathrm{or}x=2 \end{gathered}$$
Hence, −6 and 2 are the roots of the given equation.

(ii)
$$\begin{gathered} \frac{1}{2x-3}+\frac{1}{x-5}=1\frac{1}{9} \\ \Rightarrow \frac{1}{2x-3}+\frac{1}{x-5}=\frac{10}{9} \\ \Rightarrow \frac{(x-5)+(2x-3)}{(2x-3)(x-5)}=\frac{10}{9} \\ \Rightarrow \frac{3x-8}{2x^2-3x-10x+15}=\frac{10}{9} \\ \Rightarrow \frac{3x-8}{2x^2-13x+15}=\frac{10}{9} \\ \Rightarrow 27x-72=20x^2-130x+150 \\ \Rightarrow 20x^2-157x+222=0 \\ \Rightarrow 20x^2-120x-37x+222=0 \\ \Rightarrow 20x(x-6)-37(x-6)=0 \\ \Rightarrow (20x-37)(x-6)=0 \\ \Rightarrow 20x-37=0\mathrm{or}x-6=0 \\ \Rightarrow x=\frac{37}{20}\mathrm{or}x=6 \end{gathered}$$