Question

Find all the zeroes of the equation.

–3x^4+ 27x^2 + 1200 = 0

Answer 1

-12+54+1200=0

Answer 2

1254-12=0

Answer 3

that's all i have to do?

Answer 4

is it supposed to be -3x^4 +27x^2+1200=0?

Answer 5

fixed it. thanks

Answer 6

oh!!!

Answer 7

-3x^4+26x^2=-1200

Answer 8

so then what? is that all i do? this stuff is so confusing :(

Answer 9

x^2(-3x^2+27)=-1200

Answer 10

but how do i find "the zeroes" ??

Answer 11

\[-3x ^{4}+27x ^{2}+1200=0\]
\[t=x ^{2}\]
\[-3t ^{2}+27t+1200=0\]
find the zeroes of t, then substitute them in\[t=x ^{2}\]
to get the zeroes you're looking for. Don't forget its always
\[\pm \sqrt{t}\]

Answer 12

thanks!

Answer 13

no problem!