How do you solve the equation $x^{2} + 8 x - 1 = 0$ $x^2+8x-1=0$ by completing the square?

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Answer 1 · 2017-06-01T11:33:18

Solution : x~~ 0.12(2dp) or x ~~ -8.12 (2dp)#

$x^{2} + 8 x - 1 = 0 or (x^{2} + 8 x + 4^{2}) - 16 - 1 = 0$ $x^2+8x-1 = 0 or (x^2+8x +4^2) -16-1 = 0$ or

${(x + 4)}^{2} - 17 = 0 or {(x + 4)}^{2} = 17 or (x + 4) = \pm \sqrt{17}$ $(x+4)^2-17=0 or (x+4)^2 =17 or (x+4) = +- sqrt 17$ or

$:. x= -4 + sqrt17 or x= -4 - sqrt17$ or

$x~~ 0.12(2dp) or x ~~ -8.12 (2dp)$

Solution : $x~~ 0.12(2dp) or x ~~ -8.12 (2dp)$ [Ans]

How do you solve the equation x^2+8x-1=0x2+8x−1=0x^2+8x-1=0 by completing the square?