Solve each of the following quadratic equations:

Solution:

(i)

$\frac{x}{x-1}+\frac{x-1}{x}=4 \frac{1}{4}, \quad x \neq 0,1$

$\Rightarrow \frac{x^{2}+(x-1)^{2}}{x(x-1)}=\frac{17}{4}$

$\Rightarrow \frac{x^{2}+x^{2}-2 x+1}{x^{2}-x}=\frac{17}{4}$

$\Rightarrow \frac{2 x^{2}-2 x+1}{x^{2}-x}=\frac{17}{4}$

$\Rightarrow 8 x^{2}-8 x+4=17 x^{2}-17 x$

$\Rightarrow 9 x^{2}-9 x-4=0$

$\Rightarrow 9 x^{2}-12 x+3 x-4=0$

$\Rightarrow 3 x(3 x-4)+1(3 x-4)=0$

$\Rightarrow(3 x-4)(3 x+1)=0$

$\Rightarrow 3 x-4=0$ or $3 x+1=0$

$\Rightarrow x=\frac{4}{3}$ or $x=-\frac{1}{3}$

Hence, $\frac{4}{3}$ and $-\frac{1}{3}$ are the roots of the given equation.

(ii)

$\frac{x-1}{2 x+1}+\frac{2 x+1}{x-1}=2$

$\Rightarrow \frac{(x-1)^{2}+(2 x+1)^{2}}{(2 x+1)(x-1)}=2$

$\Rightarrow\left(x^{2}+1-2 x\right)+\left(4 x^{2}+1+4 x\right)=2(2 x+1)(x-1)$

$\Rightarrow 5 x^{2}+2 x+2=2\left(2 x^{2}-x-1\right)$

$\Rightarrow 5 x^{2}+2 x+2=4 x^{2}-2 x-2$

$\Rightarrow x^{2}+4 x+4=0$

$\Rightarrow x^{2}+2 x+2 x+4=0$

$\Rightarrow x(x+2)+2(x+2)=0$

$\Rightarrow(x+2)(x+2)=0$

$\Rightarrow(x+2)=0$ or $(x+2)=0$

$\Rightarrow x=-2$ or $x=-2$

$\Rightarrow x=-2$