# Solve the following quadratic equations by factorization:

Question:

Solve the following quadratic equations by factorization:

$\frac{1}{2 a+b+2 x}=\frac{1}{2 a}+\frac{1}{b}+\frac{1}{2 x}$

Solution:

$\frac{1}{2 a+b+2 x}=\frac{1}{2 a}+\frac{1}{b}+\frac{1}{2 x}$

$\Rightarrow \frac{1}{2 a+b+2 x}-\frac{1}{2 a}=\frac{1}{b}+\frac{1}{2 x}$

$\Rightarrow \frac{2 a-(2 a+b+2 x)}{(2 a+b+2 x)(2 a)}=\frac{2 x+b}{2 b x}$

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

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

$\Rightarrow\left(4 a^{2}+2 a b+4 a x\right)(2 x+b)+2 b x(2 x+b)=0$

$\Rightarrow(2 x+b)\left(4 a^{2}+2 a b+4 a x+2 b x\right)=0$

$\Rightarrow 2 x+b=0$ or $4 a^{2}+2 a b+(4 a+2 b) x=0$

$\Rightarrow x=-\frac{b}{2}$ or $x=-\frac{4 a^{2}+2 a b}{4 a+2 b}$

$\Rightarrow x=-\frac{b}{2}$ or $x=-\frac{a(4 a+2 b)}{4 a+2 b}$

$\Rightarrow x=-\frac{b}{2}$ or $x=-a$

Hence, the factors are $-a$ and $-\frac{b}{2}$.

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