How do you solve #12x ^ { 2} - 2x + 2= 4#?

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Answer 1 · 2018-03-10T18:02:27

#x_1 = 1/2#

#x_2 = -1/3#

#12x^2 - 2x + 2 -4 = 0#

#12x^2 - 2x - 2 = 0#

#"Quadratic discriminant formula ": D = b^2 - 4ac#

#D = (-2)^2 - 4*12*(-2) = 4 + 96 = 100#

#D > 0 # implies 2 solutions exist.

#x_1 = (-b + sqrt(100))/(2a) = (2 + 10)/24 = 12/24 = 1/2 #

#x_2 = (-b - sqrt(100))/(2a) = (2 -10)/ 24 = -8/24 = - 1/3#

Hence 2 roots as quaranteed by the formula :)