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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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 · 2017-05-28T14:56:35

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

Explanation:

$to complete the square$ $"to "color(blue)"complete the square"$

add ${(\frac{1}{2} coefficient of x-term)}^{2} to both sides$ $(1/2" coefficient of x-term")^2" to both sides"$

$that is add \frac{1}{2} {(\frac{4}{2})}^{2} = 4 to both sides$ $"that is add " 1/2(4/2)^2=4" to both sides"$

$x^{2} + 4 x + 4 - 1 = 0 + 4$ $x^2+4x+color(red)(4)-1=0+color(red)(4)$

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

$add 1 to both sides$ $"add 1 to both sides"$

$\Rightarrow {(x + 2)}^{2} = 5$ $rArr(x+2)^2=5$

$take the square root of both sides$ $color(blue)"take the square root of both sides"$

$\Rightarrow x + 2 = \pm \sqrt{5} \leftarrow note plus or minus$ $rArrx+2=+-sqrt5larr" note plus or minus"$

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

Answer 2 · 2017-05-28T15:07:17

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

Explanation:

$x^{2} + 4 x - 1 = 0$ $x^2+4x-1=0$
$x^{2} + 4 x + {(\frac{4}{2})}^{2} - {(\frac{4}{2})}^{2} - 1 = 0$ $x^2+4x+(4/2)^2-(4/2)^2-1=0$
${(x + 2)}^{2} - 5 = 0$ $(x+2)^2-5=0$
${(x + 2)}^{2} = 5$ $(x+2)^2=5$
$x = - 2 \pm \sqrt{5}$ $x=-2+-sqrt5$

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

2 Answers

Explanation:

Explanation:

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

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