How do you solve #6x²-81=0#?

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Answer 1 · 2015-05-23T16:53:14

We can manipulate the expression by isolating #x#. Alternatively, we can proceed to use Bhaskara.

Isolating #x#

#6x^2-81=0#
#x^2=81/6#
#x=sqrt(81/6)#
#x=+-9/sqrt(6)#
Rationalizing:
#x=+-(3sqrt(6))/6#

Using Bhaskara, where #a=6#, #b=0# and #c=-81#

#(-0+-sqrt(0-4(6)(-81)))/12#
#(+-sqrt(1944))/12#
#(+-sqrt(2^3*3^5))/12#
#(+-18sqrt(6))/12#

#+-(3sqrt(6))/2#

P.s.: there are other forms, but these seem so simple they appeal to me!