How do you solve using the completing the square method $2x^2 + 8x + 1 = 0$ ?

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How do you solve using the completing the square method $2x^2 + 8x + 1 = 0$ ?

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Answer 1 · 2016-02-27T15:14:28

$2x^2+8x+1=0$
Dividing both sides by 2 we get
$=>x^2+4x+1/2=0$
$=>x^2+2*x*2+2^2-2^2+1/2=0$
$=>x^2+2*x*2+2^2=4-1/2=7/2$
$=>(x+2)^2=7/2$
$=>x+2=+-sqrt(7/2)$
$=>x=-2+-sqrt(7/2)$

Answer 2 · 2016-02-27T15:22:37

See solution below.

Explanation:

$2(x^2 + 4x + m) = -1$

$m = (b/2)^2$

$m = (4/2)^2$

$m = 4$

$2(x^2 + 4x + 4 - 4) = -1$

$2(x + 2)^2 - 4(2) = -1$

$2(x + 2)^2 = -1 + 8$

$(x + 2)^2 = 7/2$

$x + 2 = +-sqrt(7/2)$

$x = +-sqrt(7/2) - 2$

Don't forget the $+-$ sign with the square root. Remember: a quadratic equation must always have two solutions.

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How do you solve using the completing the square method $2x^2 + 8x + 1 = 0$ ?

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How do you solve using the completing the square method $2x^2 + 8x + 1 = 0$ ?