Solve for x: 4x² + 4a²x + a⁴ + b⁴ = 0

To solve the equation 4x² + 4a²x + a⁴ + b⁴ = 0, we first need to rearrange it into a form suitable for using the quadratic formula. The equation is a quadratic in terms of x.

We can identify the coefficients in the standard quadratic form Ax² + Bx + C = 0:

  • A = 4
  • B = 4a²
  • C = a⁴ + b⁴

Next, we apply the quadratic formula:

x = (-B ± √(B² – 4AC)) / (2A)

Substituting in our coefficients, we get:

x = ( -4a² ± √((4a²)² – 4 * 4 * (a⁴ + b⁴)) ) / (2 * 4)

Calculating under the square root:

  • B² = (4a²)² = 16a⁴
  • 4AC = 4 * 4 * (a⁴ + b⁴) = 16(a⁴ + b⁴)

So, we simplify:

B² – 4AC = 16a⁴ – 16(a⁴ + b⁴) = 16a⁴ – 16a⁴ – 16b⁴ = -16b⁴

The discriminant (B² – 4AC) is negative, which means that the solutions for x will be complex numbers. Plugging this back into the formula gives:

x = (-4a² ± √(-16b⁴)) / 8

This can be further simplified as:

x = (-4a² ± 4bi) / 8

Finally, reduce:

x = (-a² ± bi) / 2

Thus, the solutions for the equation 4x² + 4a²x + a⁴ + b⁴ = 0 are:

x = (-a² + bi) / 2 and x = (-a² – bi) / 2.

More Related Questions