# How do you solve 4x^2-8x + 1 = 0?

May 11, 2018

${x}_{1} = 1 + \frac{\sqrt{3}}{2} \mathmr{and} {x}_{2} = 1 - \frac{\sqrt{3}}{2}$

#### Explanation:

$4 {x}^{2} - 8 x + 1 = 0$

4x^2-8x + 1 = 0|:4
${x}^{2} - 2 x + 1 - 1 + \frac{1}{4} = 0$
${\left(x - 1\right)}^{2} - 1 + \frac{1}{4} = 0$
${\left(x - 1\right)}^{2} - \frac{3}{4} = 0 | + \frac{3}{4}$
${\left(x - 1\right)}^{2} = \frac{3}{4} | \sqrt{}$
$x - 1 = \pm \frac{\sqrt{3}}{2} | + 1$
$x = 1 \pm \frac{\sqrt{3}}{2}$

May 11, 2018

$x = 1 \pm \frac{\sqrt{3}}{2}$ Exact values

$x \approx 0.13$ to 2 decimal places
$x \approx 1.87$ to 2 decimal places

#### Explanation:

Given: $y = 0 = 4 {x}^{2} - 8 x + 1$

Write as:
$y = 0 = 4 \left({x}^{2} - \frac{8}{4} x\right) + 1$

We now change this into the format of completing the square but in doing so we introduce a value that is not in the original equation. We 'get rid' of this by introducing a value that changes it into 0. Let this as yet unknown value be $k$

$0 = 4 \left({x}^{2} \textcolor{w h i t e}{\text{d")ubrace(-8/4)color(white)("d}} x\right) + 1$
$\textcolor{w h i t e}{\text{dddddddd.d}} \downarrow$
color(white)("ddddddd")"Halve this"
$\textcolor{w h i t e}{\text{dddddddd.d}} \downarrow$
0=color(white)("d")4(xcolor(white)("d.")obrace(-1)color(white)("d"))^2+k+1
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Set $4 {\left(- 1\right)}^{2} + k = 0 \textcolor{w h i t e}{\text{dddd")=>color(white)("dddd}} k = - 4$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

$0 = 4 {\left(x - 1\right)}^{2} - 4 + 1$

$0 = 4 {\left(x - 1\right)}^{2} - 3$

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

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

$x = 1 \pm \frac{\sqrt{3}}{2}$ Exact values

$x \approx 0.13397 \ldots$
$x \approx 1.86602 \ldots$

$x \approx 0.13$ to 2 decimal places
$x \approx 1.87$ to 2 decimal places

May 11, 2018

$x = 1 + \frac{\sqrt{3}}{2}$ or $x = 1 - \frac{\sqrt{3}}{2}$

#### Explanation:

$4 {x}^{2} - 8 x + 1 = 0$

$\therefore a {x}^{2} + b x + c = 0$

$\therefore a = 4 , b = - 8 , c = 1$

$\therefore x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$\therefore x = \frac{- \left(- 8\right) \pm \sqrt{{\left(- 8\right)}^{2} - 4 \left(4\right) \left(1\right)}}{2 \left(4\right)}$

$\therefore x = \frac{8 \pm \sqrt{48}}{8}$

$\therefore x = \frac{\sqrt{4 \cdot 4 \cdot 3}}{8}$

$\therefore x = \frac{8 \pm 4 \sqrt{3}}{8}$

$\therefore x = {\cancel{8}}^{1} / {\cancel{8}}^{1} \pm \frac{{\cancel{4}}^{1} \sqrt{3}}{\cancel{8}} ^ 2$

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

$\therefore x = 1 + \frac{\sqrt{3}}{2} \mathmr{and} x = 1 - \frac{\sqrt{3}}{2}$