Gebruik worteltrekking om vir x op te los as \( {\left( {2p + 1} \right)^2} = 16 \)

\( \begin{align} {\left( {2p + 1} \right)^2} &= 16\\ \sqrt {{{\left( {2p + 1} \right)}^2}} &= \pm \sqrt {16} \\ 2p + 1 &= \pm 4\\ 2p + 1 &= + 4 &of& &2p + 1 &= - 4\\ 2p &= 3 && &2p &= - 5\\ p &= \frac{3}{2} && &p &= - \frac{5}{2} \end{align} \)