Question

What are all the roots, both real and nonreal, of the equation x^4 + x^2 = 4x^2 + 4? Option 1: -(1/2), (1/2), -2i, 2i Option 2: -(1/2), (1/2), -1, i Option 3: -1, 1, -2i, 2i Option 4: -2, 2, -i, i

Answer 1

The roots of the equation x^4 + x^2 = 4x^2 + 4 are 2, -2, i, and -i.

Step-by-step explanation:

To find all the roots, both real and nonreal, of the equation x^4 + x^2 = 4x^2 + 4, we can start by rearranging the equation to set it equal to zero and then factor it:

• x^4 + x^2 - 4x^2 - 4 = 0
• x^4 - 3x^2 - 4 = 0

This looks like a quadratic in form, so let's substitute y = x^2, which gives us:

• y^2 - 3y - 4 = 0

Factoring the quadratic, we get:

• (y - 4)(y + 1) = 0

So the solutions for y are y = 4 and y = -1.

Substituting back for x^2, we have:

• x^2 = 4, which gives x = 2 or x = -2
• x^2 = -1, which gives x = i or x = -i (since i is the square root of -1)

Therefore, the roots are 2, -2, i, and -i