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

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Answer 1 · 2015-03-31T11:45:59

$2x^2 + 8x +1 = 0$

Reduce the coefficient of $x^2$ to $1$ by dividing all terms by $2$
#x^2+4x +1/2 = 0

Remove the constant $1/2$ from the left-side by subtracting $1/2$ from both sides of the equation.
$x^2+4x = -1/2$

To "complete the square" we are looking for a value $a$
$(x+a)^2 = x^2 +2ax +a^2$

From our equation we know that $2ax = 4x$
$rarr a=2$
and $a^2 = 4$

Add $a^2$ ( $4$ ) to both sides of the equation to "complete the square"
$x^2 + 4x +4 = 4-1/2$
or
$(x+2)^2 = 7/2$

Take the square root of both sides
$x+2$ = +-sqrt(7/2)#

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

How do you solve by completing the square: 2x^2 + 8x + 1 = 02x^2 + 8x + 1 = 0?