How do you factor #4z^{2} + 33= 24z#?

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Answer 1 · 2018-07-16T02:45:34

#color(purple)(z = 3 + sqrt3/2, color(blue)(3 - sqrt3 /2#

#4z^2 + 33 = 24x#

#4z^2 - 24 z + 33 = 0#

#"It's in the form " a x^2 + b x + c = 0#

#a = 4, b = -24, c = 33#

#"roots are " z = (-b +- sqrt(b^2 - 4 a c) / (2 a))#

#"Discriminant " d = sqrt(b^2 - 4 a c)#

#d = sqrt((-24)^2 - 4 * 4 * 33)) = sqrt(48)#

Hence both the roots are real.

#:. z = (24 +- sqrt48) / 8 = 3 +- sqrt3/2#

#color(purple)(z = 3 + sqrt3/2, color(blue)(3 - sqrt3 /2#

Answer 2 · 2018-07-16T11:30:37

#(2z-6+sqrt3)*(2z-6-sqrt3)=0#

#4z^2+33=24z#

#4z^2-24z+33=0#

#4z^2-24z+36-3=0#

#(2z-6)^2-(sqrt3)^2=0#

#(2z-6+sqrt3)*(2z-6-sqrt3)=0#