Question

Help would be appreciated(:
Find all the zeros of of the equation
-3x^4+27x^2+1200=0

Answer 1

@satellite73

Answer 2

oh there you are (:

Answer 3

start by finding the zeros of
\[-3u^2+27u+1200=0\] where \(u=x^2\)

Answer 4

best bet is to divide all by \(-3\) and start with
\[u^2-9u-400=0\]

Answer 5

by some miracle this one actually factors as
\[(u+16)(u-25)=0\] and so \(u=-16\) or \(u=25\)
now go back to \(x^2\) and write
\[x^2=-16\] or \[x^2=25\]

Answer 6

So I would have to write it like this

-3x^4+27(-16)^2+1200=0 ?

Or x^2= were my finall answers ?

Answer 7

no your job is to solve for \(x\) not rewrite it
you have
\[x^2=25\] so
\[x=\pm5\] and also
\[x^2=-16\] which is not possible with real numbers

Answer 8

Oh isee !

Thank you very much for the help! (:

Answer 9

yw

Answer 10

Thank you timo (: