Problem 25 - College Algebra Essentials

Question

Check all proposed solutions. $$ \sqrt{x-5}-\sqrt{x-8}=3 $$

Step-by-Step Solution

Answer

The original equation \( \sqrt{x-5}-\sqrt{x-8}=3 \) has no solution, since the obtained solution \( x = 7/2 \) is extraneous and does not satisfy the original equation.

Step 1: Isolate one of the square roots

Isolate the \( \sqrt{x-8} \) on one side: \( \sqrt{x-5} = \sqrt{x-8} + 3 \)

Step 2: Square both sides

Square both sides to eliminate the square root: \( (\sqrt{x-5})^2 = (\sqrt{x-8} + 3)^2 \) which simplifies to \( x - 5 = x - 16 + 6\sqrt{x-8} + 9 \)

Step 3: Further Simplification

Simplify further and isolate the \( \sqrt{x-8} \): \( 6\sqrt{x-8} - x = -14 \)

Step 4: Square both sides again

Square both sides again to get rid of the square root: \( (6\sqrt{x-8} - x)^2 = (-14)^2 \). This leads to \( 36x - 16x^2 + 36x = 196 \)

Step 5: Solve the quadratic equation

Rearrange and simplify to get a quadratic equation: \( 16x^2 - 72x + 196 = 0 \). Dividing through by 4 gives \( 4x^2 - 18x + 49 = 0 \). Factorizing the equation gives \( (2x-7)^2 = 0 \)

Step 6: Find the solutions

So the solution is \( x = 7/2 \)

Step 7: Check the solution

Substitute \( x = 7/2 \) in the original equation: \( \sqrt{7/2-5} - \sqrt{7/2-8} = -3 \) which is not equal to 3. Therefore, \( x = 7/2 \) is an extraneous root and does not satisfy the original equation.

Key Concepts

Extraneous Solutions

When solving radical equations, we often encounter what are known as extraneous solutions. These are solutions that arise during the process of manipulating equations but do not satisfy the original equation.

How do extraneous solutions occur?

• They often arise when both sides of an equation are squared, which can introduce solutions that don't work in the initial form.
• This is especially common with radical and rational equations.

It's crucial to check each solution by substituting it back into the original equation.

By re-evaluating the solution within the constraints of the initial problem, you can determine whether a solution is valid or extraneous. In our example, squaring terms introduced an extraneous solution that, when checked against the original equation, did not satisfy it.

Quadratic Equations

Quadratic equations are polynomial equations of the form \[ ax^2 + bx + c = 0 \]. In the context of our exercise, the solving process leads to a quadratic equation.

The following concepts are important when dealing with quadratic equations:

• Standard Form: Make sure the equation is in the form \[ ax^2 + bx + c = 0 \] to apply quadratic solving methods.
• Factoring: If possible, factor the equation to find the solutions. Factors must satisfy both the zero product property and the original equation.
• Quadratic Formula: If quadratics can't be factored easily, the quadratic formula is a reliable fallback:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \].

In our solution, the quadratic equation \[ 4x^2 - 18x + 49 = 0 \] resulted in a single solution after factoring, but checking it led to finding it was extraneous.

Solving Equations

Solving equations is the process of finding values (solutions) for variables that satisfy the given mathematical statement. In our problem involving radical and quadratic components, solving involves several important steps.

Here's a rough guide to solving such equations effectively:

• Isolation: Separate out the radical or expression you're interested in. This first step simplifies the equation and allows further manipulation.
• Removing Radicals: Once isolated, square both sides. Be cautious because this step can introduce extraneous solutions.
• Further Manipulation: Simplify and organize terms so the equation becomes easier to handle, possibly a quadratic form as seen in our example.

After finding potential solutions, the crucial final step is to substitute these back into the original equation. Only include solutions that work with this substitution.

This comprehensive check ensures the solution set is valid and meaningful.