Problem 20 - College Algebra with Corequisite Support

Question

For the following exercises, solve the radical equation. Be sure to check all solutions to eliminate extraneous solutions. $$ \sqrt{3 x-1}-2=0 $$

Step-by-Step Solution

Answer

The solution is \(x = \frac{5}{3}\).

Step 1: Isolate the Radical

To solve the equation \(\sqrt{3x-1} - 2 = 0\), we start by isolating the square root on one side of the equation. Add 2 to both sides to get: \(\sqrt{3x-1} = 2\).

Step 2: Eliminate the Radical

Next, eliminate the square root by squaring both sides of the equation. This gives us \((\sqrt{3x-1})^2 = 2^2\), simplifying to \(3x-1 = 4\).

Step 3: Solve for x

Now, solve for \(x\) by adding 1 to both sides to obtain \(3x = 5\), and then divide both sides by 3 to find \(x = \frac{5}{3}\).

Step 4: Check for Extraneous Solutions

Lastly, substitute \(x = \frac{5}{3}\) back into the original equation to ensure it is not extraneous: \(\sqrt{3(\frac{5}{3})-1} - 2 = 0\). Simplifying, we find \(\sqrt{5 - 1} - 2 = 0\), or \(\sqrt{4} - 2 = 0\). Since \(2 - 2 = 0\) holds true, \(x = \frac{5}{3}\) is a valid solution.

Key Concepts

Solving Equations

Solving radical equations often requires a series of steps that help simplify the equation, making it possible to isolate and determine the value of the variable.
Typically, the process begins with manipulating the equation to isolate the radical on one side. For example, we start with the equation \(\sqrt{3x-1} - 2 = 0\), and we add 2 to both sides to move the constant away from the radical. This yields \(\sqrt{3x-1} = 2\).
Once isolated, the next step is to remove or "eliminate" the radical. This is achieved by squaring both sides of the equation. Squaring the square root essentially cancels out the square root, leaving you with what's inside it. In this case, you get \(3x - 1 = 4\).
With the radical gone, solving for the variable becomes straightforward. You simply perform basic algebraic operations to solve for \(x\). In our example, this involves adding 1 to both sides and then dividing by 3, reaching a solution of \(x = \frac{5}{3}\). Remember, each step must be justified and checked as errors can arise easily.

Extraneous Solutions

When solving radical equations, checking for extraneous solutions is a crucial step. These are solutions that emerge during the solving process but don't actually satisfy the original equation.
Extraneous solutions often occur when both sides of an equation are squared. Squaring can sometimes introduce solutions that weren't present in the original problem.
To identify extraneous solutions, substitute back your solution into the initial equation. If the left-hand side equals the right-hand side of the equation, then it's a valid solution.
For instance, in the equation \(\sqrt{3x-1} - 2 = 0\), after solving for \(x = \frac{5}{3}\), we substitute back into the original equation: \(\sqrt{3(\frac{5}{3})-1} - 2 = 0\). As calculations confirm that both sides equal zero, we can be confident it is a legitimate solution and no extraneous solutions exist in this case.

Square Roots

Square roots are a fundamental concept in mathematics and often appear in radical equations. The square root of a number is a value that, when multiplied by itself, gives the original number. For example, \(\sqrt{4} = 2\) because \(2 \times 2 = 4\). Square roots can be applied in various scenarios, especially in solving equations where unknowns are under the root sign.
When dealing with square roots in equations, it is important to remember certain properties:

• Only non-negative numbers have real number square roots. The concept of square roots for negative numbers leads us into complex numbers.
• Taking the square root of a perfect square will result in a whole number.
• Squaring both sides of an equation is a common tactic to eliminate the square root symbol, simplifying the equation.

In solving radical equations like \(\sqrt{3x-1} - 2 = 0\), understanding these principles helps in correctly manipulating and solving the equation. By identifying and working with the square root, any hidden variables can be brought to light and solved in simpler, more familiar algebraic terms.