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 · 2015-07-10T07:05:43

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

Explanation:

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

Add $1$ $1$ to both sides of the equation.

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

Force a perfect square trinomial on the left side by dividing the coefficient of the $x$ $x$ term and squaring the result. Add it to both sides of the equation.

${(\frac{4}{2})}^{2} = 2^{2} = 4$ $(4/2)^2=2^2=4$

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

$x^{2} + 4 x + 4 = 5$ $x^2+4x+4=5$

We now have a perfect square trinomial on the left side in the form $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ $a^2+2ab+b^2=(a+b)^2$ , where $a = x,$ $a=x,$ and $b = 2$ $b=2$ .

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

Take the square root of both sides.

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

Subtract $2$ $2$ from both sides.

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

The two values for $x$ $x$ .

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

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

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

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Explanation:

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