Question

Solve each of the following quadratic equations:

$3\left(\frac{3x-1}{2x+3}\right)-2\left(\frac{2x+3}{3x-1}\right)=5,x\ne \frac{1}{3},-\frac{3}{2}$ [CBSE 2014]

Answer 1

$$\begin{gathered} 3\left(\frac{3x-1}{2x+3}\right)-2\left(\frac{2x+3}{3x-1}\right)=5,x\ne \frac{1}{3},-\frac{3}{2} \\ \Rightarrow \frac{3{\left(3x-1\right)}^2-2{\left(2x+3\right)}^2}{\left(2x+3\right)\left(3x-1\right)}=5 \\ \Rightarrow \frac{3\left(9x^2-6x+1\right)-2\left(4x^2+12x+9\right)}{6x^2+7x-3}=5 \\ \Rightarrow \frac{27x^2-18x+3-8x^2-24x-18}{6x^2+7x-3}=5 \end{gathered}$$
$$\begin{gathered} \Rightarrow \frac{19x^2-42x-15}{6x^2+7x-3}=5 \\ \Rightarrow 19x^2-42x-15=30x^2+35x-15 \\ \Rightarrow 11x^2+77x=0 \\ \Rightarrow 11x\left(x+7\right)=0 \end{gathered}$$
$$\begin{gathered} \Rightarrow x=0\mathrm{or}x+7=0 \\ \Rightarrow x=0\mathrm{or}x=-7 \end{gathered}$$
Hence, 0 and −7 are the roots of the given equation.