Problem 19 - Algebra 2

Question

Solve each equation. $$ \sqrt{x-6}-\sqrt{x}=3 $$

Step-by-Step Solution

Answer

There is no solution for the equation as it doesn't hold for any real number.

Step 1: Isolate one square root term

To simplify solving the equation \( \sqrt{x-6} - \sqrt{x} = 3 \), we should first isolate one of the square root terms. Add \( \sqrt{x} \) to both sides of the equation to get: \[ \sqrt{x-6} = \sqrt{x} + 3 \]

Step 2: Eliminate the square root

Next, square both sides of the equation to eliminate the square root. This gives: \[ (\sqrt{x-6})^2 = (\sqrt{x} + 3)^2 \] Simplifying both sides, we get: \[ x - 6 = x + 6\sqrt{x} + 9 \]

Step 3: Simplify the equation

Subtract \( x \) from both sides to focus on the square root term: \[ -6 = 6\sqrt{x} + 9 \] Then, subtract 9 from both sides: \[ -15 = 6\sqrt{x} \]

Step 4: Solve for the square root term

Now, divide both sides by 6 to solve for \( \sqrt{x} \): \[ \sqrt{x} = -\frac{15}{6} = -\frac{5}{2} \] However, we note that this is problematic because the square root of a real number cannot be negative.

Step 5: Analyze and conclude

Since the result \( \sqrt{x} = -\frac{5}{2} \) is not possible (as square roots of real numbers are always non-negative), check for any errors or extraneous solutions introduced by squaring. As the equation does not hold valid for any real \( x \), there is no solution.

Key Concepts

Square Roots

In mathematics, square roots are vital when solving radical equations. A square root of a number is a value that, when multiplied by itself, gives the original number. For example, the square root of 9 is 3 because \(3 \times 3 = 9\). In equation solving, square roots help in identifying variable values.
When encountering square root terms, it's essential to understand the properties they follow.

• Square roots are often denoted by the radical symbol \(\sqrt{}\).
• The square root of a non-negative number is always non-negative.
• Negative square roots lead to complex numbers, not real solutions.

In the exercise example, we deal with terms like \(\sqrt{x-6}\) and \(\sqrt{x}\). These require careful handling, especially when looking to isolate these expressions for further solving. Understanding these properties is crucial as they set the foundation for advanced algebraic solutions.

Extraneous Solutions

An extraneous solution is a solution derived from an algebraic manipulation that doesn't satisfy the original equation. They often arise in the process of solving equations involving square roots because squaring both sides can introduce solutions that were not part of the original equation. In this example, after squaring the equation \(\sqrt{x-6} = \sqrt{x} + 3\), we found \(\sqrt{x} = -\frac{5}{2}\). However, this is invalid because square roots of real numbers cannot be negative. Key things to remember about extraneous solutions:

• Always check your solutions with the original equation.
• They do not represent errors in calculations but are part of algebraic processes like squaring equations.
• Verifying solutions after achieving a potential answer is crucial.

These checks ensure that solutions conceived within the equation’s context are logically correct and truly satisfy the starting equation.

Isolation of Terms

In algebra, isolating terms means rearranging the equation so that a particular term stands alone on one side of the equation. This is a fundamental step in solving equations as it prepares the path for resolving the unknown variable.In our example, isolating serves two purposes: simplifying the solving process and preventing errors such as introducing extraneous solutions. Here's how this is done:

• Start by moving terms around utilizing basic algebraic operations, like addition or subtraction. In our problem, isolating \(\sqrt{x-6}\) involved adding \(\sqrt{x}\) to both sides.
• Focus on a single term to simplify equations step-by-step.

By strategically isolating terms, you reduce complexity and make it easier to apply further operations, like elimination of radicals or further simplification.

Solving Algebraic Equations

Solving algebraic equations involves finding values of unknowns within the equation that make it true. This often involves various algebraic strategies and manipulations. Solving radical equations like the one given often uses a methodical approach:

• Isolate terms involving radicals for clarity and simplicity.
• Perform operations like squaring both sides to eliminate radicals cautiously. Remember: each operation might introduce extraneous solutions.
• Rearrange and simplify the resulting equation to solve for unknowns effectively.

Checking the validity of the solutions is crucial. Plug back potential solutions into the original equation to ensure correctness. For example, if a mathematical step results in the solution \(\sqrt{x} = -\frac{5}{2}\), verifying such a result highlights its impossible nature, leading to more accurate conclusions.