## Intermediate Algebra (12th Edition)

$q^2-\dfrac{3}{2}qr+\dfrac{9}{16}r^2$
Using $(a+b)^2=a^2+2ab+b^2$ or the square of a binomial, the given expression, $\left( q-\dfrac{3}{4}r \right)^2 ,$ is equivalent to \begin{array}{l}\require{cancel} (q)^2+2(q)\left(-\dfrac{3}{4}r\right)+\left(-\dfrac{3}{4}r\right)^2 \\\\= q^2-\dfrac{3}{2}qr+\dfrac{9}{16}r^2 .\end{array}