Factor with quadratic formula.
In general, given \(a{x}^{2}+bx+c\), the factored form is:
\[a(x-\frac{-b+\sqrt{{b}^{2}-4ac}}{2a})(x-\frac{-b-\sqrt{{b}^{2}-4ac}}{2a})\]
In this case, \(a=16\), \(b=-1\) and \(c=\frac{1}{64}\).
\[16(x-\frac{1+\sqrt{{(-1)}^{2}-4\times 16\times \frac{1}{64}}}{2\times 16})(x-\frac{1-\sqrt{{(-1)}^{2}-4\times 16\times \frac{1}{64}}}{2\times 16})\]
Simplify.
\[16(x-\frac{1}{32})(x-\frac{1}{32})\]
Solve for \(x\).
Ask: When will \((x-\frac{1}{32})(x-\frac{1}{32})\) equal zero?
When \(x-\frac{1}{32}=0\) or \(x-\frac{1}{32}=0\)
Solve each of the 2 equations above.
\[x=\frac{1}{32}\]
