How do you solve #3n^4 - 4n^2=-1#?

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Answer 1 · 2016-04-28T12:34:43

#n=1# is one solution
After further investigation I found that:
#=>x=+-1" and "x=+-sqrt(1/3) = +-sqrt(3)/3#

Write as: #n^2(3n^2-4)=-1#

Consider #3-4=-1#

Then #n^2(3n^2-4)-=3-4=-1#

Thus #n^2(-4)-=(-4) => n=1#
Also #n^2(3n^2)-=(+3)=>n=1#

Conclusion is that #n=1#

Now it is a matter of finding the others
Tony B

Let #x^2 = X#

Then we have:
#3X^2-4X+1=0#

#=>(3X-1)(X-1)=0#

#X=1" and "1/3#

But #x^2=X#

#=>x=+-1" and "x=+-sqrt(1/3) = +-sqrt(3)/3#