How do you solve #-x^2 - 8x + 5 = 0# by completing the square?

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Answer 1 · 2016-06-27T20:47:38

#x =0.583 " or "x = -8.583#

The negative #x^2# term is not comfortable. Divide by -1.

#x^2 +8x -5 =0#

Move the constant to the other side: #" "x^2 +8x = 5#

Add on the square of half the co-efficient of x to both sides to make the square of a binomial.

#x^2 + 8x + color(red)16 = 5 + color(red)16 " "color(red)((8/2)^2#

#(x+4)^2 = 21 " square root both sides"#

#x+4 = +-sqrt21#

#x = +sqrt21 -4" or " x = -sqrt21 -4#

#x =0.583 " or "x = -8.583#