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

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Answer 1 · 2015-07-03T03:57:13

Solve y = - x^2 - 4x + 1 = 0

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

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

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

How do you solve 1−4x−x2=01 - 4x - x^2=0 using completing the square?