Question

What are the solutions of 4x^4 – 4x^2 = 8?

x = ± i

x = ±1

x= ± sqrt 2

x= ±2

Answer 1

Given: 4x^4 – 4x^2 = 8
We can solve it by rearranging the terms,

4x^4 - 4x^2 - 8 = 0
Taking 4 common from full equation, we get,
x^4 - x^2 - 2 = 0
x^4 - 2x^2 +x^2 - 2 = 0
x^2(x^2-2) + 1(x^2-2) = 0

Now,
x^2 + 1 = 0
Or x^2= -1
(Because square-root of -1 is called i)
Or x = +i or - i

Also, x^2 -2 = 0
Or x^2= 2
Or x = +√2 or -√2

Therefore, solutions are
x= ± sqrt 2 and x= ± i

Answer 2

You can solve the equation

`4x^4-4x^2=8, \\ x^4-x^2=2 \\ x^4-x^2-2=0` using the substitution
`t=x^2`.

Then the equation becomes quadratic:

`t^2-t-2=0, \\ D=(-1)^2-4\cdot (-2)\cdot 1=1+8=9, \\ √(D)=3, \\ t_(1,2) =(-(-1)\pm 3)/(2) =2,-1`.
Since
`t= x^(2) ,` you have that
`x^(2) =2, \\ x^(2) =-1=i^2.`
Solutions of these equations are:

`x_1= √(2) , \\ x_2= -√(2), \\ x_3=i, \\ x_4=-i.`
Answer: Correct choices are A and C.