Answer
the solution set is\[\left\{ 1,10 \right\}\].
Work Step by Step
Step 1:Shift all nonzero terms to left side and obtain zero on the other side.
For this, add 10 to both sides as follows:
\[{{x}^{2}}-11x+10=-10+10\].
This implies that,
\[{{x}^{2}}-11x+10=0\]
So, after shifting all nonzero terms to the left side, the above equation becomes: \[{{x}^{2}}-11x+10=0\]
Step 2:Find the factor of the above equation:
Consider the equation,\[{{x}^{2}}-11x+10=0\]
Factorize it as follows:
\[\begin{align}
& {{x}^{2}}-11x+10=0 \\
& {{x}^{2}}-10x-x+10=0 \\
& x\left( x-10 \right)-1\left( x-10 \right)=0 \\
& \left( x-1 \right)\left( x-10 \right)=0
\end{align}\]
Steps 3 and 4: Set each factor equal to zero and solve the resulting equation:
From step 2, \[\left( x-1 \right)\left( x-10 \right)=0\]
By the zero product principal, either \[\left( x-1 \right)=0\]or\[\left( x-10 \right)=0\].
Now \[\left( x-1 \right)=0\] implies that\[x=1\] and \[\left( x-10 \right)=0\]implies that\[x=10\].
Step5: Check the solution in the original equation:
Check for\[x=1\]. So consider,
\[\begin{align}
& {{x}^{2}}-11x+10=0 \\
& {{\left( 1 \right)}^{2}}-11\left( 1 \right)+10=0 \\
& 1-11+10=0 \\
& 0=0
\end{align}\]
And, check for\[x=10\].
\[\begin{align}
& {{x}^{2}}-11x+10=0 \\
& {{\left( 10 \right)}^{2}}-11\left( 10 \right)+10=0 \\
& 100-110+10=0 \\
& 0=0
\end{align}\]
Hence, the solution set is\[\left\{ 1,10 \right\}\].
