How do you solve $x^{2} - 6 x - 11 = 0$ $x^2-6x-11=0$ by completing the square?

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Answer 1 · 2016-08-13T23:43:09

$x = 3 \pm 2 \sqrt{5}$ $x = 3+-2sqrt(5)$

Move the constant term to the RHS. Take half the $x$ $x$ coefficient. Square it and add to both sides.

$x^{2} - 6 x + {(- 3)}^{2} = 11 + {(- 3)}^{2}$ $x^2 - 6x + (-3)^2 = 11 + (-3)^2$

Can simplify this to

${(x - 3)}^{2} = 20$ $(x-3)^2 = 20$

Take square roots of both sides

$x - 3 = \pm \sqrt{20} = \pm 2 \sqrt{5}$ $x - 3 = +-sqrt(20) = +-2sqrt(5)$

$x = 3 \pm 2 \sqrt{5}$ $x = 3+-2sqrt(5)$

How do you solve x^2-6x-11=0x2−6x−11=0x^2-6x-11=0 by completing the square?