Question
Solve the quadratic equation by factorization: $\frac{1}{2 a+b+2 x}=\frac{1}{2 a}+\frac{1}{b}+\frac{1}{2 x}$
✓
Answer
$\text { Consider } \frac{1}{2 a+b+2 x}=\frac{1}{2 a}+\frac{1}{b}+\frac{1}{2 x}$
$\Rightarrow \frac{1}{2 a+b+2 x}-\frac{1}{2 x}=\frac{1}{2 a}+\frac{1}{b}$
$\Rightarrow 2 ab(2 x-2 a-b-2 x)=(2 a+b) 2 x(2 a+b+2 x)$
$\Rightarrow 2 ab(-2 a-b)=2(2 a+b)\left(2 ax+bx+2 x^2\right)$
$\Rightarrow-ab=2 ax+bx+2 x^2$
$\Rightarrow 2 x^2+2 ax+bx+ab=0$
$\Rightarrow 2 x(x+a)+b(x+a)=0$
$\Rightarrow(2 x+b)(x+a)=0$
$\Rightarrow x=-a,-\frac{b}{2}$
Hence the roots are $-a,-\frac{b}{2}$.
$\Rightarrow \frac{1}{2 a+b+2 x}-\frac{1}{2 x}=\frac{1}{2 a}+\frac{1}{b}$
$\Rightarrow 2 ab(2 x-2 a-b-2 x)=(2 a+b) 2 x(2 a+b+2 x)$
$\Rightarrow 2 ab(-2 a-b)=2(2 a+b)\left(2 ax+bx+2 x^2\right)$
$\Rightarrow-ab=2 ax+bx+2 x^2$
$\Rightarrow 2 x^2+2 ax+bx+ab=0$
$\Rightarrow 2 x(x+a)+b(x+a)=0$
$\Rightarrow(2 x+b)(x+a)=0$
$\Rightarrow x=-a,-\frac{b}{2}$
Hence the roots are $-a,-\frac{b}{2}$.
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