Question

what is the solution to $-4 = -6 + \sqrt5{8x - 2}$?
a $4\frac{1}{4}$
b 4
c 8
d $\frac{3}{10}$
e $\frac{1}{2}$

Explanation:

Step1: Isolate the fifth - root term

We start with the equation \(-4=-6+\sqrt[5]{8z - 2}\). Add 6 to both sides of the equation to isolate the fifth - root term.
\(-4 + 6=-6+6+\sqrt[5]{8z - 2}\)
Simplifying the left - hand side and the right - hand side, we get \(2=\sqrt[5]{8z - 2}\)

Step2: Eliminate the fifth - root

To eliminate the fifth - root, we raise both sides of the equation to the fifth power. Recall that \((\sqrt[5]{a})^5=a\). So, if we raise both sides of \(2=\sqrt[5]{8z - 2}\) to the fifth power, we have:
\(2^5=(\sqrt[5]{8z - 2})^5\)
Since \(2^5 = 32\) and \((\sqrt[5]{8z - 2})^5=8z - 2\), the equation becomes \(32 = 8z-2\)

Step3: Solve for z

Add 2 to both sides of the equation \(32 = 8z-2\):
\(32 + 2=8z-2 + 2\)
\(34=8z\)
Then divide both sides by 8:
\(z=\frac{34}{8}=\frac{17}{4}=4\frac{1}{4}\)

Answer:

A \(4\frac{1}{4}\)