How do you solve #4x^2+1>=4x#?

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Answer 1 · 2016-07-11T01:05:36

Convert into a quadratic equation and then select test points.

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

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

#2x(2x - 1) - 1(2x - 1) = 0#

#(2x - 1)(2x - 1) = 0#

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

#x = 1/2#

#(2x - 1)^2 >= 0#

Selecting test points, we find that all values of x satisfy the inequality.

Hence, the solution is #{x|x in RR|}#.

Hopefully this helps!