How do you solve #x^2+2x+4=0# by completing the square?

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Answer 1 · 2017-01-31T04:52:13

#x=-1-isqrt3# or #x=-1+isqrt3#

#x^2+2x+4=0# can be written as

#ul(x^2+2xx x xx1 +1^2)-1^2+4=0#

and as #a^2+2ab+b^2=(a+b)^2#, this is

#(x+1)^2-1+4=0#

or #(x+1)^2+3=0#

Now, to solve the equation, we must convert it to form #a^2-b^2#, but as we have #+3#, we write it as #-(-3)#

and using imaginary numbers #-3=(isqrt3)^2#, as #i^2=-1# and #(sqrt3)^2=3#

Hence, we can write #(x+1)^2+3=0# as #(x+1)^2-(isqrt3)^2=0#

Using identity #a^2-b^2=(a+b)(a-b)#, this becomes

#(x+1+isqrt3)(x+1-isqrt3)=0#

i.e. either #x=-1-isqrt3# or #x=-1+isqrt3#.