How do you solve the equation x^2-4x+1=0 by completing the square?

Mar 4, 2018

$x = 2 \pm \sqrt{3}$

Explanation:

The difference of squares identity can be written:

${A}^{2} - {B}^{2} = \left(A - B\right) \left(A + B\right)$

We can complete the square and use this with $A = \left(x - 2\right)$ and $B = \sqrt{3}$ as follows:

$0 = {x}^{2} - 4 x + 1$

$\textcolor{w h i t e}{0} = {x}^{2} - 4 x + 4 - 3$

$\textcolor{w h i t e}{0} = {\left(x - 2\right)}^{2} - {\left(\sqrt{3}\right)}^{2}$

$\textcolor{w h i t e}{0} = \left(\left(x - 2\right) - \sqrt{3}\right) \left(\left(x - 2\right) + \sqrt{3}\right)$

$\textcolor{w h i t e}{0} = \left(x - 2 - \sqrt{3}\right) \left(x - 2 + \sqrt{3}\right)$

Hence:

$x = 2 \pm \sqrt{3}$

Mar 4, 2018

In this question you are asked to solve the given equation by the completing the square method. The whole purpose of using this method is to create a perfect square that can be factored "perfectly".

Explanation:

Right now, if you tried factoring this question it wouldn't work. So, let's complete the square!

The first step is identifying your terms, in which your current equation looks like this: $a {x}^{2} - b x + c$ = 0

For completing the square method, you will need to your 'a' and 'b' terms, and this is done by moving the 'c' term over to the other side of the equation to look like this: $a {x}^{2} - b x = c$

Now, we add in component that will lead us to completing the square, which is ${\left(\frac{b}{2}\right)}^{2}$. This means that you will add ${\left(\frac{b}{2}\right)}^{2}$ (which in this case is equal to 4), and you will have to add it to both sides of the equation to keep them balanced.

Then, once you have added that part in, solve the equation by factoring:

${x}^{2} - 4 x + 4 = 1 + 4$
${x}^{2} - 4 x + 4 = 5$
$\left(x - 2\right) \left(x - 2\right) = 5$
${\left(x - 2\right)}^{2} = 5$

Now, to find 'x' you will apply the Null Factor Law. This means you will now let 'x' = 0, and simply move the -2 over to the other side to find your answer. Note: since we are using a perfect square, x will only have one value since there is only one answer (x-2) in the brackets, meaning in this case we can ignore the exponent (as the answers will be the same). If there was more than one answer in the bracket (e.g. (x-4) and (x-6)), then we would apply Null Factor Law to both brackets and solve for x.

$\left(x - 2\right) = 5$
$x = 5 + 2$
$x = 7$