Problem 23 - College Algebra and Calculus: An Applied Approach

Question

Find the real solution(s) of the radical equation. Check your solutions. \(\sqrt{x-10}-4=0\)

Step-by-Step Solution

Answer

The real solution to the equation \(\sqrt{x-10}-4=0\) is \(x = 26\).

Step 1: Isolate the square root

Start by isolating the square root on one side of the equation. To do that, add '4' to both sides of the equation. This results in: \(\sqrt{x-10} = 4\)

Step 2: Square both sides of the equation

In order to get rid of the square root, square both sides of the equation, resulting in \((\sqrt{x-10})^2 = 4^2\). This simplifies to: \(x-10 = 16\)

Step 3: Solve for 'x'

Re-arrange the equation to solve for 'x'. \(x = 16 + 10\), which simplifies to: \(x = 26\)

Step 4: Check the solution

It is necessary to check your solution as squaring can introduce extraneous solutions. Substitute 'x' into the original equation: \(\sqrt{26-10}-4=0\), simplifying to \(\sqrt{16}-4=0\) and then to \(4-4=0\). As this is a true statement, therefore, 'x = 26' is a valid solution.

Key Concepts

Isolate the Square Root

The first step in solving radical equations is to isolate the square root on one side of the equation. Isolating the square root makes it easier to remove the radical and solve for the variable, which is crucial for finding the solution.
For instance, the equation \(\sqrt{x-10}-4=0\) presents a scenario where the square root expression is not isolated. To isolate it, you can perform simple algebraic operations such as addition or subtraction. In this example, by adding '4' to both sides of the equation, you clear the way to focus solely on the square root term. The resulting equation, \(\sqrt{x-10} = 4\), is now prepped for the next step in solving the radical equation.

• Add or subtract terms as needed to isolate the square root.
• Ensure the square root stands alone on one side of the equation for clarity.
• Prepare the equation for the squaring process that will follow.

Isolating the square root is an essential step to avoid mistakes in later stages.

Square Both Sides of the Equation

Once the square root is isolated, the next move is to eliminate the radical by squaring both sides of the equation. Squaring the isolated square root and its equivalent will reveal the root's hidden value.
When we apply this to our example, \(\sqrt{x-10} = 4\), we need to square the entire left side of the equation, \(\sqrt{x-10}\) and the entire right side, which is '4'. This step results in \(x-10 = 4^2\) or \(x-10 = 16\), since the square of 4 is 16. After squaring both sides, the equation no longer contains a radical, making it a straightforward algebraic equation.

• Ensure to square the entire side of the equation, not just the terms individually.
• Perform the squaring operation carefully to avoid introducing errors.
• Squaring is reversible only if the original value was non-negative, so this is why it's important to check for extraneous solutions.

This step brings you closer to finding the variable's value.

Check for Extraneous Solutions

After solving the squared equation for the variable, it's not yet time to celebrate! There's a final and crucial step: checking for extraneous solutions. Squaring both sides of an equation can introduce solutions that don't work in the original equation. These are extraneous, or false, solutions.
In our problem, after simplifying \(x = 16 + 10\), we found the potential solution \(x = 26\). To verify its validity, we substitute '26' back into the original equation \(\sqrt{x-10}-4=0\). After checking, \(\sqrt{26-10}-4=0\) simplifies to \(\sqrt{16}-4=0\), then to \(4-4=0\), which is true. This confirms that '26' is indeed a valid solution and not an extraneous one.

• Always substitute your solution back into the original equation.
• If the original equation holds true with your solution, then it's valid.
• If the original equation does not hold true, discard the solution as extraneous.

Verifying your solutions ensures accuracy in your final answer.