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 ):
- ( y = \frac{-2 + 6}{2} = \frac{4}{2} = 2 )
- ( y = \frac{-2 - 6}{2} = \frac{-8}{2} = -4 )
Since ( y = x^2 ), we then substitute back:
For ( y = 2 ):
[
x^2 = 2 \Rightarrow x = \pm \sqrt{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}
]
