How do you solve by completing the square for $5 - 4 x - x^{2}$ $5-4x-x^2$ ?

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How do you solve by completing the square for $5 - 4 x - x^{2}$ $5-4x-x^2$ ?

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Answer 1 · 2015-05-30T13:40:49

I assume you meant to solve $5 - 4 x - x^{2} = 0$ $5-4x-x^2 = 0$

This is equivalent to
$XXXXX$ $color(white)("XXXXX")$ $x^{2} + 4 x = 5$ $x^2+4x = 5$
Completing the square
$XXXXX$ $color(white)("XXXXX")$ $x^{2} + 4 x + 4 = 5 + 4$ $x^2+4x+4 = 5+4$
$XXXXX$ $color(white)("XXXXX")$ ${(x + 2)}^{2} = 9$ $(x+2)^2 = 9$
Take the square root
$XXXXX$ $color(white)("XXXXX")$ $x + 2 = \pm 3$ $x+2 = +-3$
Solve for x
$XXXXX$ $color(white)("XXXXX")$ $x = - 2 \pm 3$ $x=-2+-3$

$x = - 5$ $x=-5$ or $x = + 1$ $x=+1$

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How do you solve by completing the square for $5 - 4 x - x^{2}$ $5-4x-x^2$ ?

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