Problem 82 - College Algebra

Question

Solve each equation by the method of your choice. $$ \sqrt{3 x+1}-\sqrt{x-1}=2 $$

Step-by-Step Solution

Answer

The solution to the equation is \( x = 1 \)

Step 1: Isolate one of the radicals

First, move one of the square roots to the other side of the equation: \( \sqrt{3x+1} = \sqrt{x-1} + 2 \)

Step 2: Square both sides of the equation

Squaring both sides of the equation will allow removal of the square roots: \( (3x + 1) = ((x-1) + 2)^2 \) This simplifies to \(3x + 1 = x^2 - 2x + 1 + 4x - 4 + 4 \) which simplifies further to \( 3x + 1 = x^2 + 2x + 1 \)

Step 3: Simplify the equation and solve for x

The equation can be rearranged to: \( 0 = x^2 - x \) The left side of the equation factors to \( 0 = x(x - 1) \) This gives us two potential solutions, \( x = 0 \) and \( x = 1 \)

Step 4: Check the solutions

The original equation is only defined for \( x \geq 1 \) due to the square root \( \sqrt{x-1} \). Therefore, the potential solution \( x = 0 \) must be discarded. When \( x = 1 \) is substituted into the original equation, we get \( \sqrt{3(1)+1}-\sqrt{1-1}=2 \) which simplifies to \( 2 = 2 \), verifying that \( x = 1 \) is indeed a solution.

Key Concepts

Isolate Radicals

In solving radical equations, the first step is to isolate the radical expression on one side of the equation. Let's demystify this concept with an example. Suppose we're given the equation \( \sqrt{3x+1}-\sqrt{x-1}=2 \). Isolating a radical means to move it by itself on one side of the equation. Here, we add \(\sqrt{x-1}\) to both sides, resulting in \(\sqrt{3x+1} = \sqrt{x-1} + 2\). Isolating radicals simplifies your work because it sets the stage for the next step: squaring both sides, thus eliminating the square roots.

When you isolate a radical, make sure you are moving it tactfully. Any term with a radical should be moved as a whole to avoid disrupting the integrity of the expression. Furthermore, always double-check that the movement of terms doesn't introduce extraneous solutions—those false 'solutions' that might not satisfy the original equation.

Square Both Sides

After isolating the radical, the next technique we apply in solving radical equations is to 'square both sides'. This step requires a bit of finesse. For example, taking our isolated radical equation \( \sqrt{3x+1} = \sqrt{x-1} + 2 \), we raise both sides to the power of two to eliminate the square roots. Squaring \(\sqrt{3x+1}\) gives 3x+1, while squaring the right side is a bit trickier and involves expanding binomials: \((\sqrt{x-1} + 2)^2\).

This expansion will eventually help us derive a standard algebraic equation. When squaring binomials, remember to apply the distributive property accurately (also known as the FOIL method) to prevent any algebraic errors. The precision during this step is crucial as squaring affects all terms, including constants and variables.

Check Solutions

Finding potential solutions to radical equations is not the end of our journey—we must also verify them. Let's inspect the solutions to our equation, which simplified to \(x=0\) and \(x=1\). It is essential to plug these solutions back into the original equation because sometimes, we encounter solutions that do not actually satisfy the equation (extraneous solutions).

In our example, the original equation dictates that \(x\) should be equal to or larger than 1, due to the radical \(\sqrt{x-1}\). The solution \(x=0\) must be dismissed since it doesn't meet the necessary condition. When we check \(x=1\), it satisfies both sides of the equation proving to be a true solution. Checking solutions is a vital step in confirming the accuracy of your answers and should never be overlooked.

Algebraic Equations

After we have squared both sides and eliminated the radical expressions, we are left with an algebraic equation. These equations often include variables raised to a power, constants, and at times, variable products. In our example, the algebraic equation \(3x + 1 = x^2 + 2x + 1\) was derived after eliminating the radicals. The next step is simplifying and solving the equation for \(x\).

Simplification may include combining like terms and factoring. In our case, simplifying led us to \(x(x - 1) = 0\). Solving algebraic equations requires you to isolate the variable and find the value(s) of \(x\) that satisfy the equation, which sometimes involves using the zero-product property. The basic understanding of how to manipulate algebraic equations is crucial as they are the foundation of most higher-level math problems and are widely applicable in solving real-world issues.