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

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

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Answer 1 · 2015-03-31T11:45:59

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

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

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

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

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

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

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

Therefore
$x = - 2 - \sqrt{\frac{7}{2}} = - (\sqrt{2} + \sqrt{7})$ $x = -2 -sqrt(7/2) = - (sqrt(2)+sqrt(7))$
or
$x = - 2 + \sqrt{\frac{7}{2}} = \sqrt{7} - \sqrt{2}$ $x = -2 + sqrt(7/2) = sqrt(7) - sqrt(2)$

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

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