Solve for \(x\) in \({x}^{2}-4x+4=0\).
Solve for \(x\).
\[{x}^{2}-4x+4=0\]
Rewrite \({x}^{2}-4x+4\) in the form \({a}^{2}-2ab+{b}^{2}\), where \(a=x\) and \(b=2\).
\[{x}^{2}-2(x)(2)+{2}^{2}=0\]
Use Square of Difference: \({(a-b)}^{2}={a}^{2}-2ab+{b}^{2}\).
\[{(x-2)}^{2}=0\]
Take the square root of both sides.
\[x-2=0\]
Add \(2\) to both sides.
\[x=2\]
Substitute \(x=2\) into \(x+10x+25=0\).
Since \(47=0\) is not true, this is an inconsistent system.
