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

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

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Answer 1 · 2017-01-11T21:38:40

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

Explanation:

The difference of squares identity can be written:

$a_{2} - b^{2} = (a - b) (a + b)$ $a_2-b^2 = (a-b)(a+b)$

Complete the square then use this with $a = (x + 2)$ $a=(x+2)$ and $b = \sqrt{2}$ $b=sqrt(2)$ as follows:

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

$0 = x^{2} + 4 x + 4 - 2$ $color(white)(0) = x^2+4x+4-2$

$0 = {(x + 2)}^{2} - {(\sqrt{2})}^{2}$ $color(white)(0) = (x+2)^2-(sqrt(2))^2$

$0 = ((x + 2) - \sqrt{2}) ((x + 2) + \sqrt{2})$ $color(white)(0) = ((x+2)-sqrt(2))((x+2)+sqrt(2))$

$0 = (x + 2 - \sqrt{2}) (x + 2 + \sqrt{2})$ $color(white)(0) = (x+2-sqrt(2))(x+2+sqrt(2))$

Hence:

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

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

1 Answer

Explanation:

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