Rewrite \(16{x}^{2}+8x+1\) in the form \({a}^{2}+2ab+{b}^{2}\), where \(a=4x\) and \(b=1\).
\[{(4x)}^{2}+2(4x)(1)+{1}^{2}<0\]
Use Square of Sum: \({(a+b)}^{2}={a}^{2}+2ab+{b}^{2}\).
\[{(4x+1)}^{2}<0\]
Take the square root of both sides.
\[4x+1<0\]
Subtract \(1\) from both sides.
\[4x<-1\]
Divide both sides by \(4\).
\[x<-\frac{1}{4}\]
Decimal Form: -0.25
x<-1/4
