# How do you solve x^2 - 3x +2 = 0 by completing the square?

Jul 4, 2016

$x = 2 \mathmr{and} x = 1$

#### Explanation:

$a {x}^{2} + b x + c = 0$ is the general form of a quadratic trinomial.

${x}^{2} - 3 x \text{ " = -2 " move the constant to the RHS}$

Add $\textcolor{red}{{\left(\frac{b}{2}\right)}^{2}}$ to both sides to form the square of a binomial. This step is the completing of the square.

x^2 - 3x + color(red)((3/2)^2) " = -2 + color(red)((3/2)^2)

Add ${\left(\frac{b}{2}\right)}^{2}$ to both sides to form the square of a binomial.

${\left(x - \frac{3}{2}\right)}^{2} = - 2 + \frac{9}{4} = \frac{1}{4}$

$x - \frac{3}{2} = \pm \sqrt{\frac{1}{4}}$

$x = \pm \sqrt{\frac{1}{4}} + 1 \frac{1}{2}$

$x = + \frac{1}{2} + 1 \frac{1}{2} \text{ or } x = - \frac{1}{2} + 1 \frac{1}{2}$

$x = 2 \mathmr{and} x = 1$