What are the zero(s) of $f(x)=31x^4 +57 -13x^2$ ?

Question

1603 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-10-22T22:27:12

$x = +-sqrt((13+-i sqrt(6899))/62)$

$f(x) = 31x^4+57-13x^2$

$=31(x^2)^2-13(x^2)+57$

Using the quadratic formula, this has roots:

$x^2 = (13+-sqrt(13^2-(4xx31xx57)))/(2*31)$

$=(13+-sqrt(-6899))/62$

$=(13+-i sqrt(6899))/62$

So $f(x) = 0$ has roots:

$x = +-sqrt((13+-i sqrt(6899))/62)$

What are the zero(s) of f(x)=31x^4 +57 -13x^2f(x)=31x^4 +57 -13x^2?