Question

Solve each of the following quadratic equations:

$3\left(\frac{7x+1}{5x-3}\right)-4\left(\frac{5x-3}{7x+1}\right)=11,x\ne \frac{3}{5},-\frac{1}{7}$ [CBSE 2014]

Answer 1

$$\begin{gathered} 3\left(\frac{7x+1}{5x-3}\right)-4\left(\frac{5x-3}{7x+1}\right)=11,x\ne \frac{3}{5},-\frac{1}{7} \\ \Rightarrow \frac{3{\left(7x+1\right)}^2-4{\left(5x-3\right)}^2}{\left(5x-3\right)\left(7x+1\right)}=11 \\ \Rightarrow \frac{3\left(49x^2+14x+1\right)-4\left(25x^2-30x+9\right)}{35x^2-16x-3}=11 \\ \Rightarrow \frac{147x^2+42x+3-100x^2+120x-36}{35x^2-16x-3}=11 \end{gathered}$$
$$\begin{gathered} \Rightarrow \frac{47x^2+162x-33}{35x^2-16x-3}=11 \\ \Rightarrow 47x^2+162x-33=385x^2-176x-33 \\ \Rightarrow 338x^2-338x=0 \\ \Rightarrow 338x\left(x-1\right)=0 \end{gathered}$$
$$\begin{gathered} \Rightarrow x=0\mathrm{or}x-1=0 \\ \Rightarrow x=0\mathrm{or}x=1 \end{gathered}$$
Hence, 0 and 1 are the roots of the given equation.