Problem 52 - Precalculus : Building Concepts and Connections

Question

Solve the equation to find all real solutions. Check your solutions. $$x-6 \sqrt{x}=-5$$

Step-by-Step Solution

Answer

The solution to the equation \(x - 6\sqrt{x} = -5\) is \(x = 25\). The other solution \(x = 1\) is an extraneous solution introduced by squaring.

Step 1: Isolate the radical

The first step is to isolate the radical term on one side of the equation. To accomplish this, we can add \(6\sqrt{x}\) to both sides of the equation, resulting in: \(x = 6\sqrt{x} - 5\)

Step 2: Square both sides

With the radical term isolated, we can now remove the square root by squaring both sides of the equation. This gives us: \((x+5)^2 = (6\sqrt{x})^2\). Simplify to obtain: \(x^2 + 10x + 25 = 36x\)

Step 3: Rearrange to get a quadratic equation

To solve for \(x\), rearrange the equation to look like a standard quadratic equation \(ax^2 + bx + c = 0\). Subtract \(36x\) from both sides to get: \(x^2 - 26x + 25 = 0\)

Step 4: Solve the quadratic equation

Now solve the quadratic equation \(x^2 - 26x + 25 = 0\). Using the formula for roots of a quadratic equation, we obtain \(x = 1,25\)

Step 5: Check the solution

We must check these solutions in the original equation to ensure they are valid and are not extraneous solutions(which can be introduced by squaring). For \(x=1\) the equation gives us \(1 - 6*1 \neq -5\), therefore \(x=1\) is an extraneous solution. For \(x=25\) the original equation gives us \(25 - 6*5 = -5\). Therefore, \(x=25\) is our valid solution.

Key Concepts

Radical Equations

Radical equations are those that feature a variable inside a square root, cube root, or any other radical. They can appear tricky at first because we usually work with linear or quadratic equations. However, solving radical equations follows a simple set of steps.

• Isolate the Radical Term: First, try to get the term with the radical by itself on one side of the equation. This makes it simpler to eliminate the radical in the next step.
• Eliminate the Radical: If you have a square root, for example, squaring both sides of the equation will remove the square root. Be cautious here — this is the step where extraneous solutions can sneak in! More on that later.
• Complete the Solution: Once you've eliminated the radical, you will often end up with a simpler equation, such as a linear or quadratic equation, that can be solved using conventional methods.

By following these steps, you can tackle any radical equation. Just remember to verify your solutions at the end.

Quadratic Equations

Quadratic equations are polynomial equations of degree two. They follow the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants and \(x\) represents an unknown variable. These equations arise naturally when solving many types of algebraic problems.To solve quadratic equations, there are several methods:

• Factoring: This involves rewriting the quadratic as a product of two binomials. If a quadratic equation can be easily factored, it is often the quickest solution method.
• Quadratic Formula: This is a formulaic approach that works in all cases: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). It is essential to determine the discriminant \(b^2 - 4ac\) because it indicates the nature of the roots.
• Completing the Square: This method involves modifying the equation to create a perfect square trinomial, which can then be solved by taking the square root of both sides.

No matter the method you use, always check your results in the context of the original problem, especially if you transformed the equation during the process.

Extraneous Solutions

Extraneous solutions are outcomes that appear valid mathematically but do not satisfy the original equation. They can often occur when dealing with radical equations due to the squaring process, which can introduce false roots. Here's how to handle them:

• Always Verify: Once you find potential solutions, plug them back into the original equation. This ensures the solutions actually work in context.
• Watch for Radical and Absolute Expressions: Whenever you isolate radicals or square both sides of an equation, take extra care. The operations may inadvertently introduce these non-valid solutions.
• Understand Their Nature: Extraneous solutions arise because these operations can potentially alter the original nature of the equation.

By checking your solutions against the initial equation, you can confidently discard extraneous results and keep only the correct answers.