Question

Solve for x, 4underroot6x^2-13x-2underroot6=0, by completing the Square method

Answer 1

given equation is 
$4 \sqrt{6} x^2 - 13x - 2 \sqrt{6} = 0$

$4 \sqrt{6} x^2 - 16x + 3x- 2 \sqrt{6} = 0$

$( \frac{4 \sqrt{6} x}{3}+1) (3x - 2 \sqrt{6} )$

$x = \frac{2 \sqrt{6} }{3}$ and $x =\frac{-3}{4 \sqrt{6} } = \frac{ -\sqrt{6} }{8}$

Answer 2

$2 {x}^{2} - 13x + 9 = 0$

$move \: consonant \: to \: right - hand \: side \: and \: change \: its \: sign$

$2 {x}^{2} - 13x = \red{ - 9}$

$divide \: both \: side \: eq. \: by2$

$\red{ {x}^{2} - \frac{13}{2} x = \frac{ - 9}{2} }$

$add \: ( \frac{13}{4} {)}^{2} to \: both \: side \: eq.$

${x}^{2} - \frac{13x}{2} \red{+ (\frac{13}{4})^{2} } = - \frac{9}{2} \red { + ( \frac{13}{4} {)}^{2} }$

$using \: {a}^{2} - 2ab + {b}^{2} = (a - {b)}^{2} factor \: the \: eq.$

$\red{(x - \frac{13}{4} {)}^{2} } = - 9 + ( \frac{13}{4} {)}^{2}$

${(x - \frac{13}{4} {)}^{2} } = \red{ \frac{97}{16} }$

$solve \: eq. \: for \: x$

$\red{x = \frac{ - \sqrt{97} + 13}{4} }$

$\red{x = \frac{ \sqrt{97} + 13}{4} }$

$the \: eq. \: has \: 2 \: sol.$

$\boxed{ \pink{x = \frac{ - \sqrt{97} + 13}{4} }}$

${\boxed {\orange{{{x = \frac{ \sqrt{97} + 13}{4} }}}}}$