Problem 95 - Elementary and Intermediate Algebra

Question

Solve each equation. Write all proposed solutions. Cross out those that are extraneous. $$ \sqrt{6 x+2}-\sqrt{5 x+3}=0 $$

Step-by-Step Solution

Answer

The solution is \( x = 1 \), and it is not extraneous.

Step 1: Isolate One Square Root

Let's isolate one of the square roots. We can begin by moving one to the opposite side of the equation: \( \sqrt{6x + 2} = \sqrt{5x + 3} \).

Step 2: Square Both Sides

To eliminate the square roots, square both sides of the equation: \( (\sqrt{6x + 2})^2 = (\sqrt{5x + 3})^2 \). This simplifies to \( 6x + 2 = 5x + 3 \).

Step 3: Solve for x

Having a standard equation, \( 6x + 2 = 5x + 3 \), subtract \( 5x \) from both sides to obtain \( x + 2 = 3 \). Subtract 2 from both sides to solve for \( x \): \( x = 1 \).

Step 4: Verify the Solution

Substitute \( x = 1 \) back into the original equation: \( \sqrt{6(1) + 2} - \sqrt{5(1) + 3} = \sqrt{6 + 2} - \sqrt{5 + 3} = \sqrt{8} - \sqrt{8} = 0 \). This confirms that \( x = 1 \) is a valid solution.

Key Concepts

Isolating Square Roots

When solving equations involving square roots, the first primary step is to isolate one of the square roots on one side of the equation. This simplifies the equation by making it easier to manage. In the given example, \[ \sqrt{6x + 2} - \sqrt{5x + 3} = 0 \] the task is to separate one square root. We choose to move \( \sqrt{5x + 3} \) to the other side resulting in: \[ \sqrt{6x + 2} = \sqrt{5x + 3} \]. Isolating square roots means focusing on simplifying the expression, to make further steps like squaring more straightforward. This is crucial because it sets the stage for effectively eliminating square roots from the equation. Isolating a square root often involves basic algebraic manipulation such as addition, subtraction, or dividing both sides by a constant. The goal is simplicity before moving on to the next steps of solving these types of equations. Here's a simple plan on isolating square roots:

• Focus on moving terms around to get one square root by itself.
• Use basic operations like addition or subtraction to achieve isolation.
• Double-check to make sure everything else is neatly arranged for squaring next.

Remember, patience and careful manipulation are key to success in this method.

Extraneous Solutions

An extraneous solution is a solution that emerges from the algebraic process of solving an equation but doesn’t satisfy the original equation. When working with equations containing square roots, squaring both sides can introduce extraneous solutions. In our problem, squaring the isolated equations step: \[ (\sqrt{6x + 2})^2 = (\sqrt{5x + 3})^2 \] simplifies to \[ 6x + 2 = 5x + 3 \]. Solving for \( x \), we find \( x = 1 \), but does this value truly satisfy the original equation? This introduces the concept of extraneous solutions because squaring may introduce solutions that aren't meaningful or correct. It's crucial to recognize that just because a value solves the newly squared equation, it might not hold true when plugged back into the original format.How to deal with extraneous solutions:

• Always verify proposed solutions back in the original equation.
• Be aware that operations like squaring can distort the solution set.
• Eliminate any found solutions that don't meet the original equation's criteria.

Prevent unnecessary steps by always checking your definitive answers in the first version of your equation.

Verification of Solutions

Once you find a potential solution, verification is the final step. It involves plugging the solution back into the original equation to ensure it holds true. This step is crucial because it helps ascertain whether the solution found is extraneous or valid. In our problem, we have determined that \( x = 1 \) solves \( 6x + 2 = 5x + 3 \). Let's substitute back into the initial equation: \[ \sqrt{6(1) + 2} - \sqrt{5(1) + 3} = \sqrt{8} - \sqrt{8} = 0 \]. The left side simplifies to zero, confirming that \( x = 1 \) is indeed a solution that verifies correctly.Verification maintains accuracy and prevents errors associated with extraneous solutions. Here’s a quick checklist:

• Substitute the solution back into the original equation.
• Simplify both sides carefully and check they are equal.
• If they aren’t, reconsider which solutions to discard.

This process is pivotal to mathematical accuracy, enabling students to confidently arrive at verified solutions.