Solve the following equation by completing the square. 12x^2 - 48x = -39

2 answers:

7 0

Answer:

$\large\boxed{\boxed{x = \begin{cases} \frac{ \sqrt{3} }{2} + 2 \\ - \frac{ \sqrt{3} }{2} + 2 \end{cases}}}$

Step-by-step explanation:

<h3>to understand this</h3><h3>you need to know about:</h3>

<h3>let's solve:</h3>

$\sf divide \: both \: sides \: by \: 12 : \\ \sf {x}^{2} - 4x = - \frac{13}{4}$
$\sf \: add \: { - 2}^{2} \: to \: both \: sides : \\ \sf { {x}^{2} } - 4x + ( - {2}^{2} ) = \frac{13}{4} + ( { - 2}^{2} ) \\ {x}^{2} - 4x + 4 = \frac{13}{4} + 4$
$\sf simplify \: addition : \\ \sf { {x}^{2} } - 4x + 4 = \frac{3}{4}$
$\sf use \: {a}^{2} - 2ab + {b}^{2} = (a - b {)}^{2} : \\ \sf (x - 2 {)}^{2} = \frac{3}{4}$
$\sf squre \: root \: both \: sides : \\ \sf \sqrt{(x - 2 {)}^{2} } = \pm \sqrt{ \frac{3}{4} } \\ \begin{cases} x - 2 = \frac{ \sqrt{3} }{2} \\x - 2 = - \frac{ \sqrt{3} }{2} \end{cases}$
$\sf add \: 2 \: to \: both \: sides : \\ \sf \begin{cases}x = \frac{ \sqrt{3} }{2} + 2 \\ x = - \frac{ \sqrt{3} }{2} + 2 \end{cases} \\ \therefore \: x = \begin{cases} \frac{ \sqrt{3} }{2} + 2 \\ - \frac{ \sqrt{3} }{2} + 2 \end{cases}$