How do you solve $x^{2} - 4 x - 5 = 0$ $x^2 - 4x - 5 = 0$ by completing the square?

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Answer 1 · 2018-05-31T00:56:34

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$x^{2} - 4 x - 5 = 0$ $x^2-4x-5=0$

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

$x^{2} - 4 x + {(\frac{4}{2})}^{2} = 5 + {(\frac{4}{2})}^{2}$ $x^2-4x+(4/2)^2=5+(4/2)^2$

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

${(x - 2)}^{2} = 9$ $(x-2)^2=9$

$x - 2 = \pm 3$ $x-2=+-3$

$x - 2 = 3$ $x-2=3$ OR $x - 2 = - 3$ $x-2=-3$

$x - 2 = 3$ $x-2=3$
$x = 5$ $x=5$

$x - 2 = - 3$ $x-2=-3$
$x = - 1$ $x=-1$

How do you solve x2−4x−5=0x^2 - 4x - 5 = 0 by completing the square?