Question

Solve each of the following quadratic equations:

(i) $\frac{1}{x+1}+\frac{2}{x+2}=\frac{5}{x+4},x\ne -1,-2,-4$ [CBSE 2013C]
(ii) $\frac{1}{x+1}+\frac{3}{5x+1}=\frac{5}{x+4},x\ne -1,-\frac{1}{5},-4$ [CBSE 2017]

Answer 1

(i)

$$\begin{gathered} \frac{1}{x+1}+\frac{2}{x+2}=\frac{5}{x+4},x\ne -1,-2,-4 \\ \Rightarrow \frac{x+2+2x+2}{\left(x+1\right)\left(x+2\right)}=\frac{5}{x+4} \\ \Rightarrow \frac{3x+4}{x^2+3x+2}=\frac{5}{x+4} \\ \Rightarrow \left(3x+4\right)\left(x+4\right)=5\left(x^2+3x+2\right) \end{gathered}$$
$$\begin{gathered} \Rightarrow 3x^2+16x+16=5x^2+15x+10 \\ \Rightarrow 2x^2-x-6=0 \\ \Rightarrow 2x^2-4x+3x-6=0 \\ \Rightarrow 2x\left(x-2\right)+3\left(x-2\right)=0 \end{gathered}$$
$$\begin{gathered} \Rightarrow \left(x-2\right)\left(2x+3\right)=0 \\ \Rightarrow x-2=0\mathrm{or}2x+3=0 \\ \Rightarrow x=2\mathrm{or}x=-\frac{3}{2} \end{gathered}$$
Hence, 2 and $-\frac{3}{2}$ are the roots of the given equation.

(ii)

$$\begin{gathered} \frac{1}{x+1}+\frac{3}{5x+1}=\frac{5}{x+4} \\ \frac{(5x+1)+(3x+3)}{(x+1)(5x+1)}=\frac{5}{x+4} \\ (x+4)\left[(5x+1)+(3x+3)\right]=5(x+1)(5x+1) \\ (x+4)\left[8x+4\right]=25x^2+30x+5 \\ 8x^2+4x+32x+16=25x^2+30x+5 \\ 8x^2+36x+16=25x^2+30x+5 \\ 17x^2-6x-11=0 \\ 17x^2-17x+11x-11=0 \\ 17x(x-1)+11(x-1)=0 \\ (17x+11)(x-1)=0 \\ (17x+11)=0\mathrm{or}(x-1)=0 \\ x=-\frac{11}{17}\mathrm{or}x=1 \end{gathered}$$