4X⁴+8x²-32=0

To solve the equation \( 4x^4 + 8x^2 - 32 = 0 \), we can first simplify the equation by dividing all terms by 4:

\[ x^4 + 2x^2 - 8 = 0 \]

Next, we can make a substitution to make the equation easier to solve. Let \( y = x^2 \). Then, the equation becomes:

\[ y^2 + 2y - 8 = 0 \]

Now we can solve this quadratic equation using the quadratic formula:

\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

In this case, \( a = 1 \), \( b = 2 \), and \( c = -8 \):

\[ y = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-8)}}{2 \cdot 1} \]

Calculating the discriminant:

\[ y = \frac{-2 \pm \sqrt{4 + 32}}{2} \] \[ y = \frac{-2 \pm \sqrt{36}}{2} \] \[ y = \frac{-2 \pm 6}{2} \]

Now we have two possible solutions for \( y \):

1. \( y = \frac{-2 + 6}{2} = \frac{4}{2} = 2 \)
2. \( y = \frac{-2 - 6}{2} = \frac{-8}{2} = -4 \)

Since \( y = x^2 \), we then substitute back:

1. For \( y = 2 \): \[ x^2 = 2 \Rightarrow x = \pm \sqrt{2} \]

2. For \( y = -4 \): \[ x^2 = -4 \Rightarrow \text{No real solutions (since } x^2 \text{ cannot be negative)} \]

Thus, the real solutions for the original equation \( 4x^4 + 8x^2 - 32 = 0 \) are:

\[ x = \sqrt{2} \quad \text{and} \quad x = -\sqrt{2} \]