Question

Find all the zeroes of the equation.

–3x4+ 27x2 + 1200 = 0

Answer 1

-3(x^2+16)(x-5)(x+5) = 0
what are the zeros?

Answer 2

Let x^2 = y

nw
-3x4+ 27x2 + 1200 = 0



-3y^2 + 27y + 1200 = 0

nw it is a quadratic equation use quad Formula to find zeros

Answer 3

\[\frac{ -b \pm \sqrt{b^2-4ac} }{ 2a }\]

Answer 4

-3(x^4 -9x^2-400)=0
or
(x^4 - 25x^2 +16x^2 -400=0
or (x^2 -25 )(x^2+16)=0
hence (x+5)(x-5)(x+4i)(x-4i)=0
hence x=5,-5,4i,-4i (the required zeroes)
here i=sqrt(-1)