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

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Answer 1 · 2016-09-18T03:32:59

$x = - 4 \pm \sqrt{22}$ $x= -4 +- sqrt(22)$

$1 (x^{2} + 8 x + m - m) = 6$ $1(x^2 + 8x + m - m) = 6$

The goal here is to find the value of m that will make the expression in parentheses a perfect square.

$1 (x^{2} + 8 x + 16 - 16) = 6$ $1(x^2 + 8x + 16 - 16) = 6$

$1 (x^{2} + 8 x + 16) - 16 = 6$ $1(x^2 + 8x + 16) - 16 = 6$

$1 {(x + 4)}^{2} = 22$ $1(x + 4)^2 = 22$

$x + 4 = \pm \sqrt{22}$ $x + 4 = +-sqrt(22)$

$x = - 4 \pm \sqrt{22}$ $x= -4 +- sqrt(22)$

Hopefully this helps!

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