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

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Answer 1 · 2015-07-21T16:28:47

$x = \sqrt{5} - 2$ $color(green)(x = sqrt 5 - 2$ or $x = - \sqrt{5} - 2$ $color(green)(x = -sqrt 5-2$

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

To write the left hand side as a perfect square, we add 4 to both sides:

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

$x^{2} + 2 \cdot x \cdot 2 + 2^{2} = 5$ $x^2 + 2*x*2 + 2^2 = 5$

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
${(x + 2)}^{2} = 5$ $(x+2)^2 = 5$

$x + 2 = \sqrt{5}$ $x + 2 = sqrt5$ or $x + 2 = - \sqrt{5}$ $x +2 = -sqrt5$

$x = \sqrt{5} - 2$ $color(green)(x = sqrt 5 - 2$ or $x = - \sqrt{5} - 2$ $color(green)(x = -sqrt 5-2$

How do you solve the quadratic equation by completing the square: x^2+4x-1=0x2+4x−1=0x^2+4x-1=0?