Question

Solve each of the following quadratic equations:

$\frac{4}{x}-3=\frac{5}{2x+3},x\ne 0,-\frac{3}{2}$ [CBSE 2014]

Answer 1

$$\begin{gathered} \frac{4}{x}-3=\frac{5}{2x+3},x\ne 0,-\frac{3}{2} \\ \Rightarrow \frac{4}{x}-\frac{5}{2x+3}=3 \\ \Rightarrow \frac{8x+12-5x}{x\left(2x+3\right)}=3 \\ \Rightarrow \frac{3x+12}{2x^2+3x}=3 \end{gathered}$$
$$\begin{gathered} \Rightarrow \frac{x+4}{2x^2+3x}=1 \\ \Rightarrow 2x^2+3x=x+4\left(\mathrm{Cross}\mathrm{multiplication}\right) \\ \Rightarrow 2x^2+2x-4=0 \\ \Rightarrow x^2+x-2=0 \end{gathered}$$
$$\begin{gathered} \Rightarrow x^2+2x-x-2=0 \\ \Rightarrow x\left(x+2\right)-1\left(x+2\right)=0 \\ \Rightarrow \left(x+2\right)\left(x-1\right)=0 \\ \Rightarrow x+2=0\mathrm{or}x-1=0 \end{gathered}$$
$\Rightarrow x=-2\mathrm{or}x=1$
Hence, −2 and 1 are the roots of the given equation.