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

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

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Answer 1 · 2018-03-03T08:08:36

$x = \sqrt{5} + 1, - \sqrt{5} + 1 = 3.236, - 1.236$ $color(purple)(x = sqrt5 + 1, -sqrt5 + 1 = 3.236, -1.236$

Explanation:

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

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

Add 1 to both sides,

$x^{2} - 2 x + 1 = 4 + 1 = 5$ $x^2 - 2x + 1 = 4 + 1 = 5$

${(x - 1)}^{2} = {(\sqrt{5})}^{2}$ $(x-1)^2 = (sqrt5)^2$

$x - 1 = \pm \sqrt{5}$ $x-1 = +- sqrt5$

$x = \sqrt{5} + 1, - \sqrt{5} + 1 = 3.236, - 1.236$ $color(purple)(x = sqrt5 + 1, -sqrt5 + 1 = 3.236, -1.236$

Answer 2 · 2018-03-03T08:22:27

$x = 1 \pm \sqrt{5}$ $x=1+-sqrt5$

Explanation:

$using the method of completing the square$ $"using the method of "color(blue)"completing the square"$

$∙ the coefficient of the x^{2} term must be 1 which it is$ $• " the coefficient of the "x^2" term must be 1 which it is"$

$∙ add/subtract {(\frac{1}{2} coefficient of x-term)}^{2} to$ $• " add/subtract "(1/2"coefficient of x-term")^2" to"$
$x^{2} - 2 x$ $x^2-2x$

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

$\Rightarrow {(x - 1)}^{2} - 5 = 0 \leftarrow add 5 to both sides$ $rArr(x-1)^2-5=0larrcolor(blue)"add 5 to both sides"$

$\Rightarrow {(x - 1)}^{2} = 5$ $rArr(x-1)^2=5$

$Take square root of both sides$ $color(blue)"Take square root of both sides"$

$\Rightarrow x - 1 = \pm \sqrt{5} \leftarrow note plus or minus$ $rArrx-1=+-sqrt5larrcolor(blue)"note plus or minus"$

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

$\Rightarrow x = 1 \pm \sqrt{5} \leftarrow exact solutions$ $rArrx=1+-sqrt5larrcolor(red)"exact solutions"$

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

2 Answers

Explanation:

Explanation:

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Science

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Humanities

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

2 Answers

Explanation:

Explanation:

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