How do you solve $x^{2} + 4 x + 1 = 0$ $x^2+4x+1=0$ by completing the square?

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How do you solve $x^{2} + 4 x + 1 = 0$ $x^2+4x+1=0$ by completing the square?

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Answer 1 · 2016-05-06T16:24:16

$x = - 2 \pm \sqrt{3}$ $x=-2+-sqrt(3)$

Explanation:

In addition to completing the square, use the difference of squares identity:

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

with $a = (x + 2)$ $a=(x+2)$ and $b = \sqrt{3}$ $b=sqrt(3)$ as follows:

$0 = x^{2} + 4 x + 1$ $0 = x^2 + 4x + 1$

$= {(x + 2)}^{2} - 4 + 1$ $=(x+2)^2-4+1$

$= {(x + 2)}^{2} - 3$ $=(x+2)^2-3$

$= {(x + 2)}^{2} - {(\sqrt{3})}^{2}$ $=(x+2)^2-(sqrt(3))^2$

$= ((x + 2) - \sqrt{3}) ((x + 2) + \sqrt{3})$ $=((x+2)-sqrt(3))((x+2)+sqrt(3))$

$= (x + 2 - \sqrt{3}) (x + 2 + \sqrt{3})$ $=(x+2-sqrt(3))(x+2+sqrt(3))$

Hence:

$x = - 2 \pm \sqrt{3}$ $x = -2+-sqrt(3)$

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How do you solve $x^{2} + 4 x + 1 = 0$ $x^2+4x+1=0$ by completing the square?

1 Answer

Explanation:

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