# How do you solve x^2 - 6x + 4 = 0 by completing the square ?

Dec 3, 2017

Add a constant to both sides such that the left hand side is a perfect square. $x = 3 + \sqrt{5} = 3 \pm \left(2.236\right)$

#### Explanation:

In order to complete the square, we must determine what on the left hand side could be squared to give us the given ${x}^{2}$ and $x$ coefficients. A squared binomial would take the form ${\left(a x + b\right)}^{2} = {a}^{2} {x}^{2} + 2 a b x + {b}^{2}$. Looking at our equation, our $a$ is going to be 1, which means that $2 b = - 6 \to b = - 3$.

Thus, our squared binomial will take the form ${\left(x - 3\right)}^{2} = {x}^{2} - 6 x + 9$

In order to get this polynomial on the left hand side, we add 5 to both sides .

${x}^{2} - 6 x + 4 + 5 = 0 + 5 \to {x}^{2} - 6 x + 9 = 5 \rightarrow {\left(x - 3\right)}^{2} = 5$

From here, we can square root the left hand side.

$\to {\left(x - 3\right)}^{2} = 5 \to \pm \sqrt{{\left(x - 3\right)}^{2}} = \pm \sqrt{5} \to \left(x - 3\right) = \pm \sqrt{5}$

Then adding 3 to both sides:

$x = 3 \pm \sqrt{5} \approx 3 \pm 2.236 \approx 0.764 \mathmr{and} 5.236$