Solve the following quadratic equation by factorization:

$\frac{x-a}{x-b}+\frac{x-b}{x-a}=\frac{a}{b}+\frac{b}{a}$

Given:

Given quadratic equation is $\frac{x-a}{x-b}+\frac{x-b}{x-a}=\frac{a}{b}+\frac{b}{a}$.

To do:

We have to solve the given quadratic equation.

Solution:

$\frac{x-a}{x-b}+\frac{x-b}{x-a}=\frac{a}{b}+\frac{b}{a}$

$ \begin{array}{l}
\frac{( x-a)( x-a) +( x-b)( x-b)}{( x-a)( x-b)} =\frac{a( a) +b( b)}{ab}\\
\\
\frac{x^{2} -2ax+a^{2} +x^{2} -2bx+b^{2}}{x^{2} -ax-bx+ab} =\frac{a^{2} +b^{2}}{ab}\\
\\
ab\left( x^{2} -2ax+a^{2} +x^{2} -2bx+b^{2}\right) =\left( a^{2} +b^{2}\right)\left( x^{2} -ax-bx+ab\right)\\
\\
ab\left( 2x^{2} -2ax-2bx+a^{2} +b^{2}\right) =\left( a^{2} +b^{2}\right)\left( x^{2} -ax-bx+ab\right)\\
\\
2abx^{2} -2a^{2} bx-2ab^{2} x+a^{3} b+ab^{3} =a^{2} x^{2} -a^{3} x-a^{2} bx+a^{3} b+b^{2} x^{2} -ab^{2} x-b^{3} x+ab^{3}\\
\\
2abx^{2} -2a^{2} bx-2ab^{2} x=a^{2} x^{2} -a^{3} x-a^{2} bx+b^{2} x^{2} -ab^{2} x-b^{3} x\\
\\
\left( a^{2} +b^{2} -2ab\right) x^{2} +\left( -a^{3} -a^{2} b-ab^{2} -b^{3} +2a^{2} b+2ab^{2}\right) x=0\\
\\
\left( a^{2} +b^{2} -2ab\right) x^{2} +\left( -a^{3} -b^{3} +a^{2} b+ab^{2}\right) x=0\\
\\
x\left(\left( a^{2} +b^{2} -2ab\right) x+\left( -a^{3} -b^{3} +a^{2} b+ab^{2}\right)\right) =0\\
\\
x=0\ or\ \left( a^{2} +b^{2} -2ab\right) x+\left( -a^{3} -b^{3} +a^{2} b+ab^{2}\right) =0\\
\\
\left( a^{2} +b^{2} -2ab\right) x=-\left( -a^{3} -b^{3} +a^{2} b+ab^{2}\right)\\
\\
x=\frac{\left( a^{3} +b^{3} -a^{2} b-ab^{2}\right)}{\left( a^{2} +b^{2} -2ab\right)}\\
\\
x=\frac{a^{2}( a-b) +b^{2}( b-a)}{( a-b)^{2}}\\
\\
x=\frac{a^{2}( a-b) -b^{2}( a-b)}{( a-b)^{2}}\\
\\
x=\frac{\left( a^{2} -b^{2}\right)( a-b)}{( a-b)^{2}}\\
\\
x=\frac{( a+b)( a-b)( a-b)}{( a-b)^{2}}\\
\\
x=a+b\\
\\
x=0\ or\ x=a+b
\end{array}$

The values of $x$ are $0$ and $a+b$.

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Updated on: 10-Oct-2022

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