How to complete the square for this?

Question

How to express $2x^2 - 3x$ in the form $a(x+b)^2 + c$ where $a$, $b$ and $c$ are constants?

Answer 1

$$a(x+b)^2 +c= ax^2+ 2abx + ab^2+c=2x^2 -3x$$ Comparing the coefficients, $$a=2$$ $$2ab=-3$$ $$2(2)b=-3$$ $$b=\frac{-3}{4}$$ $$ab^2 +c=0$$ $$c=-ab^2=-(2)(\frac{-3}{4})^2= \frac{-9}{8}$$

Answer 2

Instead of equating coefficients, you can also follow this straightforward procedure.

First factor out the leading coefficient, so that you can work with something of the form $x^2+bx+c$, and start completing the square by writing this as $\left(x+\frac{b}2\right)^2+d$ for some $d$ that remains to be determined:

$$\begin{align*} 2x^2-3x&=2\left(x^2-\frac32x\right)\\ &=2\left[\left(x-\frac34\right)^2+d\right]\,. \end{align*}$$

Now

$$\left(x-\frac34\right)^2=x^2-\frac32x+\frac9{16}\,,$$

and what we want inside the square brackets is just $x^2-\frac32x$, so clearly $d$ must be $-\frac9{16}$ to get rid of the unwanted $\frac9{16}$ term:

$$\begin{align*} 2x^2-3x&=2\left[\left(x-\frac34\right)^2-\frac9{16}\right]\\ &=2\left(x-\frac34\right)^2-\frac98\,. \end{align*}$$