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

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Answer 1 · 2016-04-17T17:21:19

The solutions are:
$x = \sqrt{3} - 2$ $color(green)(x = sqrt 3 - 2$ or $x = - \sqrt{3} - 2$ $color(green)(x = -sqrt 3 - 2$

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

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

To write the Left Hand Side as a Perfect Square, we add 4 to both sides:

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

$x^{2} + 2 \cdot x \cdot 2 + 2^{2} = 3$ $x^2 + 2 * x * 2 + 2^2 = 3$

Using the Identity ${(a + b)}^{2} = a^{2} + 2 a b + b^{2}$ $color(blue)((a+b)^2 = a^2 + 2ab + b^2$ , we get
${(x + 2)}^{2} = 3$ $(x+2)^2 = 3$

$x + 2 = \sqrt{3}$ $x + 2 = sqrt3$ or $x + 2 = - \sqrt{3}$ $x + 2 = -sqrt3$

$x = \sqrt{3} - 2$ $color(green)(x = sqrt 3 - 2$ or $x = - \sqrt{3} - 2$ $color(green)(x = -sqrt 3 - 2$

How do you solve using the completing the square method x2+4x+1=0x^2+4x+1=0?