Question

which equation shows a valid, practical step in solving (sqrt4{2x - 8} + sqrt4{2x + 8} = 0)?
options:
(left(sqrt4{2x - 8}
ight)^4 = left(-sqrt4{2x + 8}
ight)^4)
(left(sqrt4{2x - 8}
ight)^3 = left(-sqrt4{2x + 8}
ight)^3)
(left(sqrt4{2x - 8}
ight)^4 = -left(sqrt4{2x + 8}
ight)^4)
(left(sqrt4{2x - 8}
ight)^3 = -left(sqrt4{2x + 8}
ight)^3)

Explanation:

Step1: Start with the original equation

We have the equation \(\sqrt[4]{2x - 8}+\sqrt[4]{2x + 8}=0\). Let's isolate one of the fourth - root terms. Subtract \(\sqrt[4]{2x + 8}\) from both sides of the equation:
\(\sqrt[4]{2x - 8}=-\sqrt[4]{2x + 8}\)

Step2: Raise both sides to the 4th power

To eliminate the fourth - root, we can raise both sides of the equation to the 4th power. Recall that for any real number \(a\) and \(b\), if \(a = b\), then \(a^{n}=b^{n}\) (when \(n\) is a positive integer). Here, \(n = 4\), and \(a=\sqrt[4]{2x - 8}\), \(b =-\sqrt[4]{2x + 8}\).
Raising both sides to the 4th power gives \((\sqrt[4]{2x - 8})^{4}=(-\sqrt[4]{2x + 8})^{4}\)
We know that \((-a)^{4}=a^{4}\) for any real number \(a\), so \((-\sqrt[4]{2x + 8})^{4}=(\sqrt[4]{2x + 8})^{4}\)
But we can also rewrite the right - hand side as \(-(\sqrt[4]{2x + 8})^{4}\)? Wait, no. Wait, let's check the properties again. Wait, actually, from \(\sqrt[4]{2x - 8}=-\sqrt[4]{2x + 8}\), when we raise both sides to the 4th power:
\((\sqrt[4]{2x - 8})^{4}=(-\sqrt[4]{2x + 8})^{4}\)
Since \((-y)^{4}=y^{4}\) where \(y = \sqrt[4]{2x+8}\), then \((-\sqrt[4]{2x + 8})^{4}=(\sqrt[4]{2x + 8})^{4}\), but we can also think of it as \((\sqrt[4]{2x - 8})^{4}=(-1\times\sqrt[4]{2x + 8})^{4}=(-1)^{4}\times(\sqrt[4]{2x + 8})^{4}=1\times(\sqrt[4]{2x + 8})^{4}=(\sqrt[4]{2x + 8})^{4}\)
But let's check the options. Option A is \((\sqrt[4]{2x - 8})^{4}=(-\sqrt[4]{2x + 8})^{4}\), which is equivalent to \((\sqrt[4]{2x - 8})^{4}=(\sqrt[4]{2x + 8})^{4}\) (because \((-a)^{4}=a^{4}\) for \(a=\sqrt[4]{2x + 8}\))

Answer:

The first option (the left - most option) \(\boldsymbol{(\sqrt[4]{2x - 8})^{4}=(-\sqrt[4]{2x + 8})^{4}}\) (assuming the options are ordered from left to right as the first, second, third, fourth options)