How do you use completing the square method to solve 4x^2+5x=-1?

Mar 24, 2018

$x = - \frac{1}{4}$
$x = - 1$

Explanation:

$4 {x}^{2} + 5 x = - 1$

$4 {x}^{2} + 5 x + 1 = 0$

${\left(2 x\right)}^{2} + 5 x + 1 = 0$

(2x)^2+2×(2x)×(5/4)+(5/4)^2-(5/4)^2+1=0

${\left(2 x + \frac{5}{4}\right)}^{2} - {\left(\frac{5}{4}\right)}^{2} + 1 = 0$

${\left(2 x + \frac{5}{4}\right)}^{2} = \frac{25}{16} - 1$

${\left(2 x + \frac{5}{4}\right)}^{2} = \frac{9}{16}$

$2 x + \frac{5}{4} = \pm \sqrt{\frac{9}{16}}$

$2 x + \frac{5}{4} = \frac{3}{4} \mathmr{and} 2 x + \frac{5}{4} = - \frac{3}{4}$

$2 x = \frac{3}{4} - \frac{5}{4} \mathmr{and} 2 x = - \frac{3}{4} - \frac{5}{4}$

$2 x = - \frac{2}{4} \mathmr{and} 2 x = - \frac{8}{4}$

$x = - \frac{1}{4} \mathmr{and} x = - 1$