What is the solution set for #x^2 - 2x + 5 = 0#?

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What is the solution set for #x^2 - 2x + 5 = 0#?

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Answer 1 · 2015-08-10T10:47:37

#x =(2+ sqrt( - 16))/ (2)#
#x =(2- sqrt( - 16))/ (2)#

Explanation:

Since ------- #(-2)^2# - #(4 xx1 xx 5)# < 0, x has imaginary roots

#x =(-b+- sqrt(b^2 - (4ac)))/ (2a)#
#x =(-(-2)+- sqrt((-2)^2 - (4xx1xx5)))/ (2xx1)#
#x =(-(-2)+- sqrt(4 - 20))/ (2)#
#x =(2+ sqrt( - 16))/ (2)#
#x =(2- sqrt( - 16))/ (2)#

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What is the solution set for #x^2 - 2x + 5 = 0#?

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