Solve the following quadratic equation by factorization:

$\frac{1}{(x-1)(x-2)}+\frac{1}{(x-2)(x-3)}+\frac{1}{(x-3)(x-4)}=\frac{1}{6}$

Given:

Given quadratic equation is $\frac{1}{(x-1)(x-2)}+\frac{1}{(x-2)(x-3)}+\frac{1}{(x-3)(x-4)}=\frac{1}{6}$.

To do:

We have to solve the given quadratic equation.

Solution:

$\frac{1}{(x-1)(x-2)}+\frac{1}{(x-2)(x-3)}+\frac{1}{(x-3)(x-4)}=\frac{1}{6}$

Multiplying both sides by $(x-1)(x-2)(x-3)(x-4)$, we get,

$ \begin{array}{l}
( x-1)( x-2)( x-3)( x-4)\left[\frac{1}{( x-1)( x-2)} +\frac{1}{( x-2)( x-3)} +\frac{1}{( x-3)( x-4)}\right] =( x-1)( x-2)( x-3)( x-4)\left[\frac{1}{6}\right]\\
\\
\frac{( x-1)( x-2)( x-3)( x-4)}{( x-1)( x-2)} +\frac{( x-1)( x-2)( x-3)( x-4)}{( x-2)( x-3)} +\frac{( x-1)( x-2)( x-3)( x-4)}{( x-3)( x-4)} =\frac{( x-1)( x-2)( x-3)( x-4)}{6}\\
\\
( x-3)( x-4) +( x-1)( x-4) +( x-1)( x-2) =\frac{( x-1)( x-2)( x-3)( x-4)}{6}\\
\\
6[( x-3)( x-4) +( x-1)( x-4) +( x-1)( x-2)] =( x-1)( x-2)( x-3)( x-4)\\
\\
6\left[ x^{2} -3x-4x+12+x^{2} -x-4x+4+x^{2} -x-2x+2\right] =( x-1)( x-2)( x-3)( x-4)\\
\\
6\left[ 3x^{2} -15x+18\right] =( x-1)( x-2)( x-3)( x-4)\\
\\
6\times 3\left( x^{2} -5x+6\right) =( x-1)( x-2)( x-3)( x-4)\\
\\
18\left( x^{2} -3x-2x+6\right) =( x-1)( x-2)( x-3)( x-4)\\
\\
18[ x( x-3) -2( x-3)] =( x-1)( x-2)( x-3)( x-4)\\
\\
18[( x-3)( x-2)] =( x-1)( x-2)( x-3)( x-4)\\
\\
18=( x-1)( x-4)\\
\\
x^{2} -x-4x+4=18\\
\\
x^{2} -5x+4-18=0\\
\\
x^{2} -5x-14=0\\
\\
x^{2} -7x+2x-14=0\\
\\
x( x-7) +2( x-7) =0\\
\\
( x+2)( x-7) =0\\
\\
x+2=0\ or\ x-7=0\\
\\
x=-2\ or\ x=7
\end{array}$

The values of $x$ are $-2$ and $7$.