Question

Find all the zeroes of the equation(with simple steps).

-3x^4 + 27x^2 + 1200 = 0
Please help! Urgent!

Answer 1

The zeros are,
`5,\ -5,\ 4i,\ -4i`

Solution-

`\Rightarrow -3x^4+27x^2+1200=0`

`\Rightarrow -3(x^2)^2+27(x^2)+1200=0`

Here,

a = -3, b = 27, c = 1200

So,

`x^2=(-b\pm √(b^2-4ac))/(2a)`

`=(-27\pm √(-27^2-4\cdot (-3)\cdot 1200))/(2\cdot (-3))`

`=(-27\pm √(729+14400))/(-6)`

`=(-27\pm 123)/(-6)`

`=(-27+123)/(-6),\ (-27- 123)/(-6)`

`=(96)/(-6),\ (-150)/(-6)`

`=-16,\ 25`

So,

`\Rightarrow x^2=25,\ -16`

`\Rightarrow x=√(25),\ √(-16)`

`\Rightarrow x=5,\ -5,\ 4i,\ -4i`