How do you solve #2x ^ { 2} - 2x + 7= 5#?

1 Answer
Rajat Chatterjee
Nov 3, 2016

ans.# x=(1+isqrt3)/2# or# (1-isqrt3)/2#.

Explanation:

#2x^2-2x+7=5#
or,#2x^2-2x+2=0#
# or,x^2-x+1=0# # (i)#
now according to the rule if #ax^2+bx+c=0# then #x=(-b+sqrt(b^2-4ac))/(2a)# or# (-b-sqrt(b^2-4ac))/(2a)#
hence from the equation #(i)# we get the ans.#x=(1+isqrt3)/2# or #(1-isqrt3)/2#.
here #i=sqrt(-1)#.

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