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Solve the following quadratic equation by factorization:
$\frac{a}{x-a}+\frac{b}{x-b}=\frac{2c}{x-c}$
Given:
Given quadratic equation is $\frac{a}{x-a}+\frac{b}{x-b}=\frac{2c}{x-c}$.
To do:
We have to solve the given quadratic equation.
Solution:
$\frac{a}{x-a}+\frac{b}{x-b}=\frac{2c}{x-c}$
Multiplying both sides by $(x-a)(x-b)(x-c)$, we get,
$ \begin{array}{l}
( x-a)( x-b)( x-c)\left[\frac{a}{x-a} +\frac{b}{x-b}\right] =( x-a)( x-b)( x-c)\left[\frac{2c}{x-c}\right]\\
\\
\frac{a( x-a)( x-b)( x-c)}{x-a} +\frac{b( x-a)( x-b)( x-c)}{x-b} =\frac{2c( x-a)( x-b)( x-c)}{x-c}\\
\\
a\left( x^{2} -bx-cx+bc\right) +b\left( x^{2} -ax-cx+ac\right) =2c\left( x^{2} -ax-bx+ab\right)\\
\\
ax^{2} -abx-acx+abc+bx^{2} -abx-bcx+abc=2cx^{2} -2acx-2bcx+2abc\\
\\
( a+b-2c) x^{2} +( -ab-ac-ab-bc+2ac+2bc) x+2abc-2abc=0\\
\\
( a+b-2c) x^{2} +( -2ab+ac+bc) x=0\\
\\
x[( a+b-2c) x+( -2ab+ac+bc)] =0\\
\\
x=0\ or\ ( a+b-2c) x+( -2ab+ac+bc) =0\\
\\
( a+b-2c) x=2ab-ac-bc\\
\\
x=\frac{2ab-ac-bc}{a+b-2c}\\
\\
x=0\ or\ x=\frac{2ab-ac-bc}{a+b-2c}\\
\end{array}$
The values of $x$ are $0$ and $\frac{2ab-ac-bc}{a+b-2c}$.