Question

Find all the zeroes of the equation.

-3x^4+27x^2+1200=0

Answer 1

@Vuriffy

Answer 2

y=x^2

Answer 3

-3y^2+27y+1200=0
-3(y^2-9y-400)=0
y^2-9y-400=0
can u factor this, and find solution for y

Answer 4

(y-25)(y+16)=0

Answer 5

and since y=x^2
\((x^2-25)(x^2+16)=0\)

from there, we can set each term equal to 0 and solve
\(x^2-25=0\\x^2=25\\x=\sqrt25\\x=-5,5\)


\((x^2+16)=0
\\x^2=-16
\\x=\sqrt{-16}
\\x=4i,-4i
\)

Answer 6

Note that the given equation, -3x^4+27x^2+1200=0, can be factored by dividing each of the four terms by -3. Try doing this. I agree it's a good approach to let x^2 = y.
Having factored out -3, and subst. y for x^2, what do you get?

Answer 7

\(-3(y^2)- 9y- 400)= 0 \)
\(y^2)- 9y- 400= 0 \)

@mathmale