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 $x^2=X$

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

How do you solve 3n^4 - 4n^2=-13n^4 - 4n^2=-1?