Problem 20 - College Algebra

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 Expression

To solve the given equation, we first isolate the square root. The equation is \( \sqrt{3x - 1} - 2 = 0 \). To isolate the square root, add 2 to both sides of the equation: \( \sqrt{3x - 1} = 2 \).

Step 2: Eliminate the Square Root

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

Step 3: Solve for x

Now that we have the equation \(3x - 1 = 4\), solve for \(x\). Add 1 to both sides to get \(3x = 5\), then divide by 3 to find \(x = \frac{5}{3}\).

Step 4: Check the Solution for Extraneous Solutions

To verify that \(x = \frac{5}{3}\) is a valid solution, substitute it back into the original equation: \(\sqrt{3(\frac{5}{3}) - 1} - 2 = 0\). This simplifies to \(\sqrt{5 - 1} - 2 = 0 \), or \(\sqrt{4} - 2 = 0\), which results in \(2 - 2 = 0\). Since this is true, \(x = \frac{5}{3}\) is a valid solution.

Key Concepts

Isolating Radical Expressions

When tackling a radical equation, the first clear step is to isolate the radical expression. In our exercise, the equation is \( \sqrt{3x - 1} - 2 = 0 \). The term that contains the square root is the focal point here, \( \sqrt{3x - 1} \). To effectively isolate this part, we need to eliminate any constants on the same side. In this case, your task is to get rid of \(-2\) by adding 2 to both sides, resulting in \( \sqrt{3x - 1} = 2 \). This step is essential because you need the radical on its own to address it directly in further steps. By isolating the radical, we set up the equation for the next phase without any other terms complicating our calculations.

Eliminating Square Roots

Once the radical is isolated, our next goal is to eliminate the square root to solve for \( x \) more directly. This comes from squaring both sides of the equation. For \( \sqrt{3x - 1} = 2 \), squaring both sides results in \((\sqrt{3x - 1})^2 = 2^2\). On the left, the square and the square root cancel each other out, simplifying to \( 3x - 1 \). On the right, we have \( 4 \). So, our simplified equation is \( 3x - 1 = 4 \).

• Remember, squaring is a powerful tool. It undoes the square root, but also it can introduce extraneous solutions, which we'll discuss soon.
• This simplification allows us to look at the equation without radical expressions, making it easier to solve for the variable \( x \).

Since the radical is gone, solving for \( x \) is just basic algebra from here on out.

Checking for Extraneous Solutions

Extraneous solutions can sneak in when you square both sides to eliminate a square root. Such solutions might satisfy the squared version of the equation but not the original radical equation because squaring is not reversible if any integers involved are negative.

• To check for these, substitute the found solution back into the original equation.
• For \( x = \frac{5}{3} \), plug it back as \( \sqrt{3(\frac{5}{3}) - 1} - 2 = 0 \).
• Calculate: This simplifies to \( \sqrt{5 - 1} - 2 = 0 \).
• Simplify further: \( \sqrt{4} - 2 = 0 \) turns into \( 2 - 2 = 0 \), confirming our solution is indeed correct.

It's crucial to always verify, ensuring no extraneous solutions exist, especially when eliminating radicals through squaring.