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

Sep 12, 2017

For ${x}^{2} - 3 x + 3 = 0$

Divide through by the coefficient of ${x}^{2}$ if this is not $1$:

Move the constant to the right hand side.

${x}^{2} - 3 x = - 3$

Add to both sides the square of half the coefficient of $x$

The square of half the coefficient in this case is ${\left(\frac{3}{2}\right)}^{2}$

So we have: ${x}^{2} - 3 x + {\left(\frac{3}{2}\right)}^{2} = - 3 + {\left(\frac{3}{2}\right)}^{2}$

Convert ${x}^{2} - 3 x + {\left(\frac{3}{2}\right)}^{2}$ into the square of a binomial.

${\left(x - \frac{3}{2}\right)}^{2}$

We now have : ${\left(x - \frac{3}{2}\right)}^{2} = - 3 + {\left(\frac{3}{2}\right)}^{2}$

Collect terms on right hand side:

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

Taking roots of both sides:

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

$\sqrt{\left(\frac{1}{4} \cdot 3 \cdot - 1\right)} \implies \sqrt{\frac{1}{4}} \cdot \sqrt{3} \cdot \sqrt{- 1}$

$\sqrt{- 1} = \textcolor{b l u e}{i}$
where $i$ is the imaginary unit.

$x = \frac{3}{2} \pm \left(\sqrt{\frac{1}{4}} \cdot \sqrt{3} \cdot \sqrt{- 1}\right)$

taking any exact roots and simplifying:

$x = \frac{3}{2} \pm \left(\frac{1}{2}\right) \sqrt{3} \textcolor{b l u e}{i}$

Note:
If this is your first time solving quadratics by completion of the square, it would have been best to choose an equation that had real rather than imaginary roots. This would have been easier to start of with.