How do you solve the quadratic with complex numbers given #2x^2-6x+7=0#?

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Answer 1 · 2016-09-30T11:17:45

#x=3/2+isqrt5/2# or #x=3/2-isqrt5/2#

#2x^2-6x+7=0# - we first ty to make complete square as in #(x+a)^2#

#2x^2-6x+7=0#

#hArrx^2-3x+7/2=0#

or #x^2-2xx3/2xx x+(3/2)^2-(3/2)^2+7/2=0#

or #(x-3/2)^2-9/4+14/4=0#

or #(x-3/2)^2+5/4=0#

or #(x-3/2)^2-(sqrt5/2xxi)^2=0#

As this is of the form #a^2-b^2=(a+b)(a-b)#

The above is equal to

#(x-3/2-isqrt5/2)(x-3/2+isqrt5/2)0#

Hence #x-3/2-isqrt5/2=0# or #x-3/2+isqrt5/2=0#

i.e. #x=3/2+isqrt5/2# or #x=3/2-isqrt5/2#