(x^2+3)^2=4x^2+12

2 min read Jun 17, 2024
(x^2+3)^2=4x^2+12

Solving the Equation: (x^2 + 3)^2 = 4x^2 + 12

This article will explore the solution to the equation (x^2 + 3)^2 = 4x^2 + 12. We'll break down the steps to solve this equation and find the possible values for x.

Expanding and Simplifying

First, we expand the left side of the equation using the formula (a + b)^2 = a^2 + 2ab + b^2:

(x^2 + 3)^2 = x^4 + 6x^2 + 9

Now, we have:

x^4 + 6x^2 + 9 = 4x^2 + 12

Subtracting 4x^2 + 12 from both sides gives:

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

Factoring the Equation

This equation is now a quadratic equation in terms of x^2. We can factor it as follows:

(x^2 + 3)(x^2 - 1) = 0

Finding the Solutions

Now, we have two factors that equal zero. This means either one or both of them must be equal to zero:

  • x^2 + 3 = 0
    This equation has no real solutions, as the square of any real number is non-negative.
  • x^2 - 1 = 0 Adding 1 to both sides gives: x^2 = 1 Taking the square root of both sides (remembering both positive and negative roots): x = ±1

Conclusion

Therefore, the solutions to the equation (x^2 + 3)^2 = 4x^2 + 12 are x = 1 and x = -1.