How do you solve by completing the square: $x^{2} + 4 x + 2 = 0$ $x^2 + 4x + 2 = 0$ ?

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How do you solve by completing the square: $x^{2} + 4 x + 2 = 0$ $x^2 + 4x + 2 = 0$ ?

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Answer 1 · 2015-03-30T01:31:09

x^2+4x+2=0#

Remove the $2$ $2$ from the left side expression by subtracting $2$ $2$ from both sides:
$x^{2} + 4 x = - 2$ $x^2+4x=-2$

If the left side is to be a square its non- $x$ $x$ term must be the coefficient of $x$ $x$ divided by $2$ $2$ then squared (i..e.#(4/2)^2 = 4 $); a d d t h i s a m o u n t \to ⊥ h s i d e s :$ $); add this amount to both sides:$ x^2+4x+4=+2#

Rewrite the left-hand side as a square:
${(x + 2)}^{2} = 2$ $(x+2)^2 = 2$

Take the square root of both sides
$x + 2 = \pm \sqrt{2}$ $x+2 = +-sqrt(2)$

Isolate $x$ $x$ on the rights side (by subtracting $2$ $2$ from both sides)
$x = - 2 \pm \sqrt{2}$ $x = -2+-sqrt(2)$

So the solutions are
$x = - 2 + \sqrt{2}$ $x=-2+sqrt(2)$ and $x = - 2 - \sqrt{2}$ $x=-2-sqrt(2)$

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How do you solve by completing the square: $x^{2} + 4 x + 2 = 0$ $x^2 + 4x + 2 = 0$ ?

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