# How do you solve 3x^2 - 5x+4=0 by completing the square?

May 10, 2018

$3 {\left(x - \frac{5}{6}\right)}^{2} + \frac{23}{12}$

#### Explanation:

$3 {x}^{2} - 5 x + 4$
$3 \left({x}^{2} - \frac{5}{3} x + \frac{4}{3}\right)$
$3 \left({\left(x - \frac{5}{6}\right)}^{2} - {\left(\frac{5}{6}\right)}^{2} + \frac{4}{3}\right)$
$3 {\left(x - \frac{5}{6}\right)}^{2} - 3 {\left(\frac{5}{6}\right)}^{2} + 4$
$3 {\left(x - \frac{5}{6}\right)}^{2} - \frac{25}{12} + 4$
$3 {\left(x - \frac{5}{6}\right)}^{2} + \frac{23}{12}$

I double checked and worked backwards and gives you the original function. I knew I was working with fractions the minute I saw the $3 {x}^{2}$ which I didn't mind, as long as you prioritise the $3 {x}^{2}$ and turn that into ${x}^{2}$ by itself so that completing the square would be easier.

May 10, 2018

$x = \frac{5}{6} \pm \frac{\sqrt{23}}{6} i$

#### Explanation:

Write as:

$3 \left({x}^{2} - \frac{5}{3} x\right) + 4 = 0 \text{ } \ldots \ldots \ldots \ldots \ldots . E q u a t i o n \left(1\right)$

The next step introduces a value that is not in the original equation. We get rid of it by introducing the the as yet unknown value $k$. We will use this to turn the introduced value into 0

$\textcolor{g r e e n}{\textcolor{w h i t e}{\text{dd.d}} 3 \left({x}^{2} - \frac{5}{3} x\right) + 4 = 0}$
$\textcolor{w h i t e}{\text{ddddd.dddd}} \downarrow$

$\text{halve the } - \frac{5}{3}$

$\textcolor{w h i t e}{\text{ddddd.dddd}} \downarrow$
color(green)(color(white)("ddd.d")3(x-5/6)^2+k+4=0)" ".............Equation(2)
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
The introduced value is $3 {\left(- \frac{5}{6}\right)}^{2}$ so:

Set: $3 {\left(- \frac{5}{6}\right)}^{2} + k = 0$ giving

$3 \left(+ \frac{25}{36}\right) + k = 0$

25/12+k=0color(white)("d")=>color(white)("d")k=-25/12" "......Equation(3)
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Substitute $E q n \left(3\right)$ into $E q n \left(2\right)$

$3 {\left(x - \frac{5}{6}\right)}^{2} - \frac{25}{12} + 4 = 0$

$3 {\left(x - \frac{5}{6}\right)}^{2} + \frac{23}{12} = 0$

${\left(x - \frac{5}{6}\right)}^{2} = - \frac{23}{36}$

$x - \frac{5}{6} = \pm \sqrt{- \frac{23}{36}}$

$x = \frac{5}{6} \pm \frac{\sqrt{23}}{6} i$