Solve each of the following quadratic equations:

Solution:

(i)

$\frac{1}{x+1}+\frac{2}{x+2}=\frac{5}{x+4}, \quad x \neq-1,-2,-4$

$\Rightarrow \frac{x+2+2 x+2}{(x+1)(x+2)}=\frac{5}{x+4}$

$\Rightarrow \frac{3 x+4}{x^{2}+3 x+2}=\frac{5}{x+4}$

$\Rightarrow(3 x+4)(x+4)=5\left(x^{2}+3 x+2\right)$

$\Rightarrow 3 x^{2}+16 x+16=5 x^{2}+15 x+10$

$\Rightarrow 2 x^{2}-x-6=0$

$\Rightarrow 2 x^{2}-4 x+3 x-6=0$

$\Rightarrow 2 x(x-2)+3(x-2)=0$

$\Rightarrow(x-2)(2 x+3)=0$

$\Rightarrow x-2=0$ or $2 x+3=0$

$\Rightarrow x=2$ or $x=-\frac{3}{2}$

Hence, 2 and $-\frac{3}{2}$ are the roots of the given equation.

(ii)

$\frac{1}{x+1}+\frac{3}{5 x+1}=\frac{5}{x+4}$

$\frac{(5 x+1)+(3 x+3)}{(x+1)(5 x+1)}=\frac{5}{x+4}$

$(x+4)[(5 x+1)+(3 x+3)]=5(x+1)(5 x+1)$

$(x+4)[8 x+4]=25 x^{2}+30 x+5$

$8 x^{2}+4 x+32 x+16=25 x^{2}+30 x+5$

$8 x^{2}+36 x+16=25 x^{2}+30 x+5$

$17 x^{2}-6 x-11=0$

$17 x^{2}-17 x+11 x-11=0$

$17 x(x-1)+11(x-1)=0$

$(17 x+11)(x-1)=0$

$(17 x+11)=0$ or $(x-1)=0$

$x=-\frac{11}{17}$ or $x=1$