# How do you solve using the completing the square method 4t^2 = 8t - 1?

Mar 14, 2016

See details below to obtain
$\textcolor{w h i t e}{\text{XXX}} t = 1 \pm \frac{\sqrt{3}}{2}$

#### Explanation:

Given
$\textcolor{w h i t e}{\text{XXX}} 4 {t}^{2} = 8 t - 1$

Get all terms involving the variable on the left side
$\textcolor{w h i t e}{\text{XXX}} 4 {t}^{2} - 8 t = - 1$

Divide both sides by $4$
$\textcolor{w h i t e}{\text{XXX}} {t}^{2} - 2 t = - \frac{1}{4}$

If ${t}^{2} - 2 t$ are the first two terms of an expanded squared binomial,
then the third term must be $1$.
Complete the square by adding $1$ to both sides:
$\textcolor{w h i t e}{\text{XXX}} {t}^{2} - 2 t + 1 = \frac{3}{4}$

Re-write the left side as a squared binomial
$\textcolor{w h i t e}{\text{XXX}} {\left(t - 1\right)}^{2} = \frac{3}{4}$

Taking the square roots:
$\textcolor{w h i t e}{\text{XXX}} t - 1 = \pm \frac{\sqrt{3}}{2}$

Add $1$ to both sides:
$\textcolor{w h i t e}{\text{XXX}} t = 1 - \frac{\sqrt{3}}{2} \mathmr{and} t = 1 + \frac{\sqrt{3}}{2}$

Mar 14, 2016

$t = 1 \pm \frac{\sqrt{3}}{2}$

#### Explanation:

Move all values with $t$ to the left hand side.

$4 {t}^{2} - 8 t = - 1$

Divide everything by the initial coefficient $4$ to make ${t}^{2}$ have a coefficient of 1

${t}^{2} - 2 t = - \frac{1}{4}$. Note $b = - 2$

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

${t}^{2} - 2 t + {\left(- \frac{2}{2}\right)}^{2} = - \frac{1}{4} + {\left(- \frac{2}{2}\right)}^{2}$

Rewrite in perfect square form and simplify the right hand side

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

Take $\pm$ the square root of both sides

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

Add one to both sides

$t = 1 \pm \frac{\sqrt{3}}{2}$