How do you solve the quadratic using the quadratic formula given #3d^2-5d+6=0# over the set of complex numbers?

1 Answer
Alan P.
Oct 11, 2016

#d=5/6+sqrt(47)/6icolor(white)("XX")orcolor(white)("XX")d=5/6-sqrt(47)/6i#

Explanation:

For a parabolic equation of the form:
#color(white)("XXX")color(red)ad^2+color(blue)bd+color(green)c=0#
the quadratic formula gives the solution as
#color(white)("XXX")d=(-color(blue)b+-sqrt(color(blue)b^2+4color(red)acolor(green)c))/(2color(red)a)#
#color(white)("XXXXXXXXXX")#over the set of complex numbers

For the given example:
#color(white)("XXX")color(red)a=color(red)3#
#color(white)("XXX")color(blue)b=color(blue)(-5)#
#color(white)("XXX")color(green)c=color(green)6#

Therefore
#color(white)("XXX")d=(-color(blue)(""(-5))+-sqrt(color(blue)(""(-5))^2-4xxcolor(red)3xxcolor(green)6))/(2xxcolor(red)3)#

#color(white)("XXX")=(5+-sqrt(-47))/6#

#d=5/6+sqrt(47)/6icolor(white)("XX")orcolor(white)("XX")d=5/6-sqrt(47)/6i#

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