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 · 2015-04-03T01:05:31

(Please note that I edited your question to place a plus sign between the $4 x$ $4x$ and the $2$ $2$ ; if you meant something else, perhaps a minus sign, please re-post).

$x^{2} - 4 x + 2 = 0$ $x^2-4x+2=0$

Subtract the constant ( $2$ $2$ ) from both sides
$x^{2} - 4 x = - 2$ $x^2-4x=-2$

If $x^{2} + 4 x$ $x^2 +4x$ are the first two terms of a square of the form ${(x + a)}^{2}$ $(x+a)^2$
then
$a = - 2$ $a=-2$ and the term needed to complete the square is $a^{2} = 4$ $a^2 =4$

Add 4 to both sides of the equation
$x^{2} - 4 x + 4 = 4 - 2$ $x^2-4x+4 = 4-2$

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

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

Giving the solutions
$x = 2 \pm \sqrt{2}$ $x = 2+-sqrt(2)$

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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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