How do you solve $x^{2} - 8 x - 6 = 0$ $x^2 - 8x - 6=0$ using completing the square?

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Answer 1 · 2015-07-23T02:43:35

First you change the form into $x^{2} - 8 x = 6$ $x^2-8x=6$
(add $6$ $6$ to both sides)

Method: you halve the $x$ $x$ -coefficient ( $- 8$ $-8$ ) to get $- 4$ $-4$ . The square of this is to be added to both sides.

$\to x^{2} - 8 x + 16 = 6 + 16$ $->x^2-8x+16=6+16$

${(x - 4)}^{2} = 22 \to x - 4 = \pm \sqrt{22}$ $(x-4)^2=22->x-4=+-sqrt22$

$x_{1, 2} = 4 \pm \sqrt{22}$ $x_(1,2)=4+-sqrt22$ (two solutions)
graph{x^2-8x-6 [-52.02, 52.07, -26, 26]}

How do you solve x2−8x−6=0x^2 - 8x - 6=0 using completing the square?