Question

Help please anyone?
Find all the zeroes of the equation.
-3x^4 + 27x^2 + 1200 = 0

Answer 1

anyone know how to do this one?

Answer 2

x=5

Answer 3

the answer is 5

Answer 4

can you explain to me how you got that answer?

Answer 5

I don't know how to do the steps

Answer 6

its a biquadratic ecuation, to find the zeros first we are going to call:\[x ^{2}=u\] so:\[y=-3u ^{2}+27u+1200\] now you can use bhaskara to find the zeros:\[u _{1},u_2=\frac{ -b \pm \sqrt{b ^{2}-4.a.c} }{ 2a }=\frac{ -27\pm \sqrt{27^{2}-4.(-3).1200} }{ 2.(-3)}\] and the solutions are\[u _{1}=-16\]\[u_2=25\]But:\[u _{1}=-16=x _{1}^{2}\rightarrow x _{1}=\pm \sqrt{-16}=\pm 4i\]\[u _{2}=25=x _{2}^{2}\rightarrow x _{2}=\pm \sqrt{25}= \pm5\]Finally the 4solutions to the biquadratic ecuations are:\[x _{1}=5, x _{2}=-5,x _{3}=-4i,x_4=4i\]