How do you solve #x^2+6x-16=0#?

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Answer 1 · 2016-03-27T00:06:26

#x=2# or #x=-8#

Find two factors of #16# which differ by #6#. The pair #8, 2# works in that #8 xx 2 = 16# and #8 - 2 = 6#.

Hence we find:

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

So #x = 2# or #x = -8#

#color(white)()#
Alternatively, complete the square then use the difference of squares identity:

#a^2-b^2 = (a-b)(a+b)#

with #a = (x+3)# and #b=5# as follows:

#0 = x^2+6x-16#

#=(x+3)^2-9-16#

#=(x+3)^2-25#

#=(x+3)^2-5^2#

#=((x+3)-5)((x+3)+5)#

#=(x-2)(x+8)#

Hence zeros #x=2# and #x=-8#