## Introductory Algebra for College Students (7th Edition)

$x^2+\dfrac{1}{2}x+\dfrac{1}{16} \Rightarrow\left( x+\dfrac{1}{4} \right)^2$
The third term of a perfect square trinomial is equal to the square of half the coefficient of the middle term. Hence, to complete the square of the given expression $x^2+\dfrac{1}{2}x ,$ the third term must be \begin{array}{l}\require{cancel}\left( \dfrac{1/2}{2}\right)^2 \\\\= \left( \dfrac{1}{4}\right)^2 \\\\= \dfrac{1}{16} .\end{array} Using $a^2\pm2ab+b^2=(a\pm b)^2$, then \begin{array}{l}\require{cancel} x^2+\dfrac{1}{2}x+\dfrac{1}{16} \Rightarrow\left( x+\dfrac{1}{4} \right)^2 .\end{array}