Chapter 6.7, Problem 41PPS

To determine

To Solve:

  ${\left(4z-1\right)}^{\frac{1}{5}}-1=2$

Answer to Problem 41PPS

  $z=61$

Explanation of Solution

Given Information:

Given equation:

  ${\left(4z-1\right)}^{\frac{1}{5}}-1=2$

Concept Used :

Solve Radical equations:

• Isolate the radical on one side of the equation.
• Raise each side of the equation to a power equal to the index of the radical to eliminate the radical .
• Solve the resulting polynomial equation.Check your results.
• When solving radical equations, the result may be a number that does not satisfy the original equation . Such a number is called an extraneous solution.

Calculation:

Given ${\left(4z-1\right)}^{\frac{1}{5}}-1=2$

$\begin{array}{l}{\left(4z-1\right)}^{\frac{1}{5}}-1=2\\ \Rightarrow {\left(4z-1\right)}^{\frac{1}{5}}=2+\mathrm{1...............}[Add\text{ 1 each side]}\\ \Rightarrow {\left[{\left(4z-1\right)}^{\frac{1}{5}}\right]}^5=3^5\mathrm{...............}[\text{Raise each side to the power 5=index]}\\ \Rightarrow 4z-1=243\\ \Rightarrow 4z=243+\mathrm{1.........................}[Add\text{ 1 each side]}\\ \Rightarrow 4z=244\\ \Rightarrow \frac{4z}{4}=\frac{244}{4}\mathrm{...........................}[\text{Divide each side by 4]}\\ \Rightarrow z=61\end{array}$

So, the solution is $z=61$