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

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Answer 1 · 2016-09-18T03:32:59

#x= -4 +- sqrt(22)#

#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 + 8x + 16 - 16) = 6#

#1(x^2 + 8x + 16) - 16 = 6#

#1(x + 4)^2 = 22#

#x + 4 = +-sqrt(22)#

#x= -4 +- sqrt(22)#

Hopefully this helps!