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

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Answer 1 · 2018-03-24T06:05:02

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

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

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

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

${(2 x)}^{2} + 2 \times (2 x) \times (\frac{5}{4}) + {(\frac{5}{4})}^{2} - {(\frac{5}{4})}^{2} + 1 = 0$ $(2x)^2+2×(2x)×(5/4)+(5/4)^2-(5/4)^2+1=0$

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

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

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

$2 x + \frac{5}{4} = \pm \sqrt{\frac{9}{16}}$ $2x+5/4=+-sqrt(9/16)$

$2 x + \frac{5}{4} = \frac{3}{4} or 2 x + \frac{5}{4} = - \frac{3}{4}$ $2x+5/4=3/4 or 2x+5/4=-3/4$

$2 x = \frac{3}{4} - \frac{5}{4} or 2 x = - \frac{3}{4} - \frac{5}{4}$ $2x=3/4-5/4 or 2x= -3/4-5/4$

$2 x = - \frac{2}{4} or 2 x = - \frac{8}{4}$ $2x=-2/4 or 2x= -8/4$

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

How do you use completing the square method to solve 4x^2+5x=-14x2+5x=−14x^2+5x=-1?