How do you solve #x^2+8x+1=0# by completing the square?

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Answer 1 · 2016-06-06T08:51:56

#x=-4-sqrt15# or

#x=-4+sqrt15#

In #x^2+8x+1=0#, we observe that coefficient of #x# is #8# and hence, we can complete the square by adding square of half of #8# i.e. #4^2=16#.

Hence #x^2+8x+1=0# can be written as #x^2+8x+16-16+1=0# or

#(x+4)^2-15=0# which can be written as

#(x+4)^2-(sqrt15)^2=0#

Now using identity #a^2-b^2=(a+b)(a-b)#, this can be written as

#(x+4+sqrt15)(x+4-sqrt15)=0#

Hence either #x+4+sqrt15=0# i.e. #x=-4-sqrt15#

or #x+4-sqrt15=0# i.e. #x=-4+sqrt15#