How do you solve $x^{2} + 6 x + 2 = 0$ $x^2 + 6x + 2 = 0$ using completing the square?

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How do you solve $x^{2} + 6 x + 2 = 0$ $x^2 + 6x + 2 = 0$ using completing the square?

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Answer 1 · 2015-06-11T10:20:39

A square is allways of the form $x^{2} + 2 a x + a^{2}$ $x^2+2ax+a^2$

Explanation:

Completing the square is you take half of the number $2 a$ $2a$ with x and square it, which would be $x^{2} + 6 x + 9$ $x^2+6x+9$ , but to even out with the $2$ $2$ you've got, you write $2 = 9 - 7$ $2=9−7$ :
$x^{2} + 6 x + 9 - 7 = 0 \to$ $x^2+6x+9−7=0→$
${(x + 3)}^{2} - 7 = 0 \to {(x + 3)}^{2} = 7 \to$ $(x+3)^2−7=0→(x+3)^2=7→$
$x + 3 = \pm \sqrt 7 \to x_{12} = - 3 \pm \sqrt 7$ $x+3=±√7→x_12=−3±√7$

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How do you solve $x^{2} + 6 x + 2 = 0$ $x^2 + 6x + 2 = 0$ using completing the square?

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How do you solve $x^{2} + 6 x + 2 = 0$ $x^2 + 6x + 2 = 0$ using completing the square?