How do you solve #6x^2+72=0#?

1 Answer
Alan P.
Oct 27, 2016

There are no value Real solutions to this equation.
Using Complex values #x=+-2sqrt(3)i#

Explanation:

Given
#color(white)("XXX")6x^2+72=0#

Dividing both sides by #6# gives
#color(white)("XXX")x^2+12 = 0#

#color(white)("XXX")x^2= -12#

...for all Real values of #x# we know that #x^2>=0#
so there can be no Real valued solutions.

#color(white)("XXX")x^2=(2sqrt(3))^2xx(-1)#

#color(white)("XXX")x =+-2sqrt(3)xx sqrt(-1)#

#color(white)("XXX")x= +-2sqrt(3)i#

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