How do you solve $x^{2} - 8 x - 1 = 8$ $x^2-8x-1=8$ by completing the square?

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Answer 1 · 2018-06-15T14:30:19

$x^{2} - 8 x - 1 = 8$ $x^2-8x-1=8$

$x^{2} - 8 x - 9 = 0$ $x^2-8x-9=0$

${(x - 4)}^{2} - 16 - 9 = 0$ $(x-4)^2-16-9=0$

${(x - 4)}^{2} - 25 = 0$ $(x-4)^2-25=0$

${(x - 4)}^{2} = 25$ $(x-4)^2=25$

$(x - 4) = \pm 5$ $(x-4)=\pm5$

$x - 4 = 5 or x - 4 = - 5$ $x-4=5 or x-4=-5$

$x = 9 or x = - 1$ $x=9 or x=-1$

How do you solve x2−8x−1=8x^2-8x-1=8 by completing the square?