How do you solve the equation #x^2-3/2x-23/16=0# by completing the square?

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Answer 1 · 2016-11-13T16:56:09

#x=3/4+sqrt2# or #x=3/4-sqrt2#

#x^2-3/2x-23/16=0#

completing square of form #(x-a)^2=x^2-2ax+a^2#, above becomes

#x^2-2xx3/4xx x+(3/4)^2-(3/4)^2-23/16=0#

or #(x-3/4)^2-9/16-23/16=0# or #(x-3/4)^2-32/16=0#

or #(x-3/4)^2-(sqrt2)^2=0#

and factorizing it using #a^2-b^2=(a+b)(a-b)#, we get

#(x-3/4-sqrt2)(x-3/4+sqrt2)=0#

Hence, either #x-3/4-sqrt2=0# i.e. #x=3/4+sqrt2#

or #x-3/4+sqrt2=0# i.e. #x=3/4-sqrt2#