Problem 16 - Algebra and Trigonometry

Question

Solve each radical equation in Exercises 11–30. Check all proposed solutions. $$\sqrt{6 x+1}=x-1$$

Step-by-Step Solution

Answer

The solution to the equation \(\sqrt{6x + 1} = x - 1\) is an empty set.

Step 1: Isolate the Radical Expression

The equation \(\sqrt{6x + 1} = x - 1\) already has the radical expression isolated on one side, which means there's no further rearrangement needed.

Step 2: Square Both Sides of Equation

To eliminate the square root, square both sides of the equation. This gives: \((\sqrt{6x + 1})^2 = (x - 1)^2\), which simplifies to: \(6x + 1 = x^2 - 2x + 1\). Rearrange this to get a quadratic equation: \(x^2 - 8x = 0\).

Step 3: Solve the Quadratic Equation

Factor the quadratic equation to find the potential solutions: \(x(x - 8) = 0\). The solutions are \(x = 0\) and \(x = 8\).

Step 4: Check the Proposed Solutions

Substitute the potential solutions back into the original equation to verify if they are actual solutions. For \(x = 0\), \(\sqrt{6(0) + 1} = 0 - 1 \Rightarrow 1 \neq -1\), so \(x = 0\) is not a solution. For \(x = 8\), \(\sqrt{6(8) + 1} = 8 - 1 \Rightarrow 5 = 7\), so \(x = 8\) is also not valid. This equation has no solution. Hence, the solution set is empty.

Key Concepts

Isolate the Radical Expression

When solving equations involving radicals, the initial step is often to make sure the radical expression is on one side of the equation by itself. This forms the foundation for simplifying complex equations and working towards a solution. In our example, \(\sqrt{6x + 1} = x - 1\), the radical expression \(\sqrt{6x + 1}\) is already isolated, allowing us to proceed to the next step without the need for further manipulation.
Isolating the radical is critical as it paves the way for us to apply algebraic rules, such as exponentiation, to both sides of the equation without interference from other terms.

Square Both Sides of the Equation

To deal with radical expressions, we can eliminate them by raising both sides of the equation to a power that matches the root. For square roots, we square both sides. Applying this step to \(\sqrt{6x + 1}\), we get the equation \(\sqrt{6x + 1})^2 = (x - 1)^2\), which simplifies nicely, removing the radical and enabling us to transition toward a more familiar quadratic equation.
Squaring the radical and the other side is a delicate operation though, as it can introduce extraneous solutions—solutions that do not satisfy the original equation.

Solve the Quadratic Equation

Once we have a quadratic equation, such as \(x^2 - 8x = 0\), we proceed to solve it using techniques like factoring, completing the square, or applying the quadratic formula. For our equation, factoring is the most straightforward method, resulting in \(x(x - 8) = 0\).
The Zero Product Property states that if a product equals zero, then at least one of the factors must be zero. This tells us our potential solutions are \(x = 0\) and \(x = 8\). However, we’re not done yet; we must verify these solutions in the context of the original equation.

Check the Proposed Solutions

Our last step involves substituting the potential solutions into the original radical equation to check their validity. This step is crucial because squaring both sides of the equation might introduce false solutions that do not satisfy the original equation. For our exercise, upon checking \(x = 0\) and \(x = 8\), neither value resulted in equal expressions on both sides of the equation, therefore they are both rejected as solutions.
Verifying solutions protects us from mistakenly accepting extraneous solutions, ensuring the integrity of our final answer. In this case, we conclude that the solution set is empty, as neither candidate solves the original radical equation.