Question

Find all the zeroes of the equation.

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

Answer 1

Answer is below..........

Answer 2

The zeros of the equation
`-3x^4 + 27x^2 + 1200 = 0` are 4i,-4i,5,-5.

Explanation:

Given : Equation
`-3x^4 + 27x^2 + 1200 = 0`

To find : All the zeroes of the equation ?

Solution :

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

First we divide whole equation by 3,

`-x^4 +9x^2 +400 = 0`

Now, Let
`x^2=y`

So,
`-y^2 +9y+400 = 0`

Solve by quadratic formula,
`y=(-b\pm√(b^2-4ac))/(2a)`

Here, a=-1, b=9 and c=400

`y=(-9\pm√(9^2-4(-1)(400)))/(2(-1))`

`y=(-9\pm√(1681))/(-2)`

`y=(-9\pm 41)/(-2)`

`y=(-9+41)/(-2),(-9-41)/(-2)`

`y=-16,25`

Substituting back,

`x=\pm√(y)`

When y=-16,

`x=\pm√(-16)`

`x=\pm 4i`

When y=25,

`x=\pm√(25)`

`x=\pm 5`

The zeros of the equation
`-3x^4 + 27x^2 + 1200 = 0` are 4i,-4i,5,-5.