Question

challenge problem: 3\sqrt4{2x - 7} + 2 = 5

Explanation:

Step1: Isolate the radical term

Subtract 2 from both sides of the equation \(3\sqrt[4]{2x - 7}+2 = 5\).
We get \(3\sqrt[4]{2x - 7}=5 - 2=3\).

Step2: Solve for the radical

Divide both sides by 3: \(\sqrt[4]{2x - 7}=\frac{3}{3} = 1\).

Step3: Eliminate the fourth - root

Raise both sides to the fourth power. Since \((\sqrt[4]{a})^4=a\) for \(a\geq0\), we have \((\sqrt[4]{2x - 7})^4=1^4\).
This simplifies to \(2x-7 = 1\).

Step4: Solve for x

Add 7 to both sides: \(2x=1 + 7=8\).
Divide both sides by 2: \(x=\frac{8}{2}=4\).

Step5: Check the solution

Substitute \(x = 4\) into the original equation:
Left - hand side: \(3\sqrt[4]{2(4)-7}+2=3\sqrt[4]{8 - 7}+2=3\sqrt[4]{1}+2=3\times1 + 2=5\).
Right - hand side: 5.
Since the left - hand side equals the right - hand side, \(x = 4\) is a valid solution.

Answer:

\(x = 4\)