## Precalculus (10th Edition)

$\dfrac{9}{4}$
We complete the square: $x^2-3x=x^2-2\cdot\dfrac{3}{2}x=x^2-2\cdot\dfrac{3}{2}x+\left(\dfrac{3}{2}\right)^2-\left(\dfrac{3}{2}\right)^2$ $=\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{4}$ To complete the square $x^2-3x$, add $\dfrac{9}{4}$.