Problem 21 - College Algebra with Corequisite Support

Question

For the following exercises, solve the radical equation. Be sure to check all solutions to eliminate extraneous solutions. $$ \sqrt{x-7}=5 $$

Step-by-Step Solution

Answer

The solution to the equation is \( x = 32 \).

Step 1: Remove the Radical

To remove the square root, square both sides of the equation. This yields: \( (\sqrt{x-7})^2 = 5^2 \) Therefore, \( x - 7 = 25 \)

Step 2: Solve for x

Add 7 to both sides of the equation to isolate \( x \).\( x - 7 + 7 = 25 + 7 \) This simplifies to: \( x = 32 \)

Step 3: Check for Extraneous Solutions

Substitute \( x = 32 \) back into the original equation to check it:\( \sqrt{32 - 7} = 5 \) Simplifying:\( \sqrt{25} = 5 \) Since this is true, \( x = 32 \) is a valid solution.

Key Concepts

Extraneous Solutions

When solving radical equations, checking for extraneous solutions is crucial. These are solutions that might emerge during the manipulation of the equation but do not actually satisfy the original equation. This often happens due to the algebraic manipulations, like squaring both sides, which can introduce false solutions.

To avoid falling into the trap of extraneous solutions, always substitute your solution back into the original equation. Only the solutions that hold true under these original conditions can be considered valid. In our exercise, for example, after solving we found \( x = 32 \), and substituting gives us \( \sqrt{32-7} = 5 \). This holds as true, confirming that \( x = 32 \) is not extraneous. Extraneous solutions can complicate problem-solving, so verification is a vital final step.

• Always recheck your solution with the initial equation.
• Be aware that manipulation may introduce false roots.
• Confirmation of solutions is necessary to avoid incorrect conclusions.

Square Roots

Square roots are a fundamental concept in solving radical equations. The square root of a number \( x \) is a value that, when multiplied by itself, yields \( x \). Mathematically, if \( y = \sqrt{x} \), then \( y^2 = x \). Understanding this allows us to remove radicals from equations efficiently.

In our problem, we have \( \sqrt{x-7}=5 \). We eliminate the radical by squaring both sides: \( (\sqrt{x-7})^2 = 5^2 \). This fundamental step simplifies to \( x-7 = 25 \), making it easier to solve for \( x \).
Basic properties of square roots include:

• The principal square root is always non-negative when dealing with real numbers.
• Square roots are undone by squaring, a powerful tool for simplification.

Understanding these properties allows us to tackle radical equations with confidence and accuracy.

Step-by-Step Solutions

Solving equations step-by-step ensures thorough understanding and precision. This structured method means tackling each component one at a time, ensuring no detail clusters.

For the equation \( \sqrt{x-7}=5 \):

• **Step 1: Remove the Radical.** Square both sides to simplify the equation: \((\sqrt{x-7})^2 = 5^2\), giving \(x-7 = 25\).
• **Step 2: Solve for x.** Isolate \(x\) by adding 7 to both sides, resulting in \(x=32\).
• **Step 3: Check for Extraneous Solutions.** Substitute \(x=32\) back into the original equation to verify: \(\sqrt{32-7}=5\), confirming the solution is correct.

Following a structured, step-by-step approach prevents errors and ensures all outcomes are investigated thoroughly. It provides a systematic path from given problem to solution, promoting clarity and comprehension for any variety of equations.