(x^2+1)^2-5x^2-5=0

2 min read Jun 17, 2024

Solving the Equation: (x^2 + 1)^2 - 5x^2 - 5 = 0

This equation may look intimidating at first glance, but we can solve it by using algebraic manipulation and a bit of creativity. Here's how:

Start by expanding the squared term:

(x^2 + 1)^2 = (x^2 + 1)(x^2 + 1) = x^4 + 2x^2 + 1

Now, substitute this back into the original equation:

x^4 + 2x^2 + 1 - 5x^2 - 5 = 0

Combine like terms:

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

This equation can be factored by recognizing it as a quadratic in x^2:

(x^2 - 4)(x^2 + 1) = 0

Now, we have two factors, and for the product to equal zero, at least one of the factors must equal zero.

Case 1: x^2 - 4 = 0

Solving for x^2:

x^2 = 4

Taking the square root of both sides:

x = ±2

Case 2: x^2 + 1 = 0

Solving for x^2:

x^2 = -1

Since the square of any real number cannot be negative, there are no real solutions for this case.

Therefore, the solutions to the equation (x^2 + 1)^2 - 5x^2 - 5 = 0 are x = 2 and x = -2.