- Cube both sides of the equation \(\sqrt[3]{2x + 5} = 5\) to eliminate the radical, resulting in \(2x + 5 = 125\).
- Isolate the variable term by subtracting 5 from both sides, yielding \(2x = 120\).
- Divide by 2 to find the final value of \(x = 60\).
- The solution is \(\boxed{60}\).
Explanation
- Problem Analysis
The goal is to solve for \(x\) in the equation \(\sqrt[3]{2x + 5} = 5\). This is an algebraic equation involving a cube root, which we can solve by isolating the variable \(x\). The first step is to eliminate the radical by using the inverse operation of taking the cube root, which is raising both sides of the equation to the power of 3.
- Cubing Both Sides
By cubing both sides of the equation \(\sqrt[3]{2x + 5} = 5\), we obtain:
\[ (\sqrt[3]{2x + 5})^3 = 5^3 \]
Since the cube of a cube root returns the original expression inside the radical, the left side simplifies to \(2x + 5\). On the right side, calculating \(5^3\) gives \(5 \times 5 \times 5 = 125\). Thus, our equation becomes:
\[ 2x + 5 = 125 \]
- Solving for x
Now we isolate the term with \(x\) by subtracting 5 from both sides of the equation:
\[ 2x = 125 - 5 \]
\[ 2x = 120 \]
Finally, we solve for \(x\) by dividing both sides by 2:
\[ x = \frac{120}{2} = 60 \]
The solution to the equation is \(x = 60\). Comparing this result to the given options, we find that it matches the option \(60\).
Examples
Solving equations with radicals is fundamental in fields like physics and engineering. For example, if you are determining the physical dimensions of an object based on its volume where the relationship is defined by a cubic root, such as the side length of a cube \(s\) given its volume \(V\) (\(s = \sqrt[3]{V}\)), solving such an equation allows you to find the exact dimensions needed for construction or design. Mastery of this skill ensures accuracy in technical calculations.