4x4 + 8x2 - 12 = 0
To solve the equation \(4x^4 + 8x^2 - 12 = 0\), we can first simplify it.
Divide the entire equation by 4 to make calculations easier:
\[ x^4 + 2x^2 - 3 = 0 \]
Substitution: Let \(y = x^2\). Then, the equation becomes:
\[ y^2 + 2y - 3 = 0 \]
Factor the quadratic equation:
We need two numbers that multiply to \(-3\) and add up to \(2\). Those numbers are \(3\) and \(-1\).
Therefore, we can factor the equation as:
\[ (y + 3)(y - 1) = 0 \]
Find the values of \(y\):
Setting each factor to zero gives us:
\[ y + 3 = 0 \quad \Rightarrow \quad y = -3 \]
\[ y - 1 = 0 \quad \Rightarrow \quad y = 1 \]
Substitute back for \(x\):
Since \(y = x^2\):
For \(y = -3\): \[ x^2 = -3 \quad \Rightarrow \quad x = \pm i\sqrt{3} \]
For \(y = 1\): \[ x^2 = 1 \quad \Rightarrow \quad x = \pm 1 \]
Final Solutions:
Thus, the solutions for the equation \(4x^4 + 8x^2 - 12 = 0\) are:
\[ x = 1, \quad x = -1, \quad x = i\sqrt{3}, \quad x = -i\sqrt{3} \]
