# Solve each of the following quadratic equations:

Question:

Solve each of the following quadratic equations:

$3\left(\frac{7 x+1}{5 x-3}\right)-4\left(\frac{5 x-3}{7 x+1}\right)=11, \quad x \neq \frac{3}{5}, \quad-\frac{1}{7}$

Solution:

$3\left(\frac{7 x+1}{5 x-3}\right)-4\left(\frac{5 x-3}{7 x+1}\right)=11, \quad x \neq \frac{3}{5},-\frac{1}{7}$

$\Rightarrow \frac{3(7 x+1)^{2}-4(5 x-3)^{2}}{(5 x-3)(7 x+1)}=11$

$\Rightarrow \frac{3\left(49 x^{2}+14 x+1\right)-4\left(25 x^{2}-30 x+9\right)}{35 x^{2}-16 x-3}=11$

$\Rightarrow \frac{147 x^{2}+42 x+3-100 x^{2}+120 x-36}{35 x^{2}-16 x-3}=11$

$\Rightarrow \frac{47 x^{2}+162 x-33}{35 x^{2}-16 x-3}=11$

$\Rightarrow 47 x^{2}+162 x-33=385 x^{2}-176 x-33$

$\Rightarrow 338 x^{2}-338 x=0$

$\Rightarrow 338 x(x-1)=0$

$\Rightarrow x=0$ or $x-1=0$

$\Rightarrow x=0$ or $x=1$

Hence, 0 and 1 are the roots of the given equation.