Question

Find all zeros: f(x)=x^4+9x^2+20

I know you're supposed to set it equal to zero and solve, but I can't think at all right now. I can't figure out how to solve x^4+9x^2+20=0!!! I'm totally screwed for tomorrow's final. Please Help! >.<

Answer 1

This polynomial has no zero because :
\[\forall x\in{\mathbb R},\quad x^4+9x^2+20>0\]

Answer 2

oh, the answer is supposed to be i root 5, -i root 5, 2i, -2i

Answer 3

\[x={-b\pm\sqrt{b^2-4ac}\over2a}\]where \(ax^2+bx+c=0\)

Answer 4

Rewrite it like so:$$f(x)=(x^2)^2+9(x^2)+20$$hmm... isn't that a quadratic in \(x^2\)? What if we replaced \(x^2\) with something like \(u\)?$$f(x)=u^2+9u+20$$Try setting that equal to \(0\) and factoring:$$u^2+9u+20=0\\u^2+5u+4u+20=0\\u(u+5)+4(u+5)=0\\(u+4)(u+5)=0$$We know that for the product to be \(0\) either must be \(0\) as well:$$u+4=0\quad\implies u=-4\\u+5=0\quad\implies u=-5$$But wait, these aren't values of \(x\), they're of \(u\). We need to map them backwards. Recall we let \(u=x^2\):$$x^2=-4\\x^2=-5$$Stop. If you only want *real* roots, there's nothing left to do here -- we know that no real numbers have negative squares. If you want complex roots, we can keep going, however:$$x^2=-4\implies x=\pm\sqrt{-4}=\pm2i\\x^2=-5\implies x=\pm\sqrt{-5}=\pm i\sqrt5$$

Answer 5

... and now we have all of our roots! Notice that because it's a 4th degree polynomial, it yields 4 complex (in our case, pure imaginary) roots. Noura was correct in noting there were no *real* roots (\(\forall x\in\mathbb{R}\) means 'for all real numbers \(x\)'), but there are indeed imaginary ones.

Answer 6

ooh, I totally remember doing that now. Thank you so much!