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How do you solve # x^2-9=0#?

↳Redirected from "How do you evaluate #sin^-1 (-sqrt3/2)#?"

1 Answer

Mar 4, 2018

#x=+-3#

Explanation:

#=>x^2-9=0#
#=>x^2=9#
#=>x=sqrt9#

#=>x=sqrt (3^2)#
#=>x=3#

#=>x=sqrt ((-3)^2)#
#=>x=-3#

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