Please complete the square:(2x^2) + 8x + 1 = 0?

Answer 1

2x² + 8x + 1 = 0
2(x² + 4x ) = -1
2(x² + 4x + 4) = -1 + 8
2(x + 2)² = 7
(x + 2)² = 7/2 = 14/4
x + 2 = ±(√14)/2
x = -2 ±(√14)/2

Answer 2

2x² + 8x + 1 = 0
You may do this in this way too. You need to remember this formula;

x= -b ± √ b² -4a * c / 2a


where equation is in the form of ax²+bx+c=0


Now just substitute values form your equation to this formula:

Your answer will be:
x= -8 ± √ 64 – 4*2*1 / 2*2

i.e. x= -2 ± √56 / 4 or x = -2 ± √14*4/4

or x = -2 ± √14/ 2

Answer 3

2x^2 + 8x + 1 = 0
multiply by 2
4x^2 + 16x + 2 = 0
2x +2(2x)(4) + 16 - 16 + 2 = 0
(2x + 4)^2 - 14 = 0
(2x + 4)^2 - (sqrt(14))^2 = 0
[2x + 4 +sqrt(14)][2x + 4 - sqrt(14)] = 0

Answer 4

2(x^2 + 4x) + 1 = 0
now you need to add four inside the parentheses and subtract 8 outside of it.
2(x^2 + 4x + 4) + 1 - 2(4)
2(x+2)^2 - 7 = 0
2(x+2)^2 = 7
(x+2)^2 = 7/2
x+2 = squareRoot(7/2)
x = -2 + squareRoot(7/2)
x = -2 - squareRoot(7/2)