Problem 25 - College Algebra

Question

For the following exercises, solve the radical equation. Be sure to check all solutions to eliminate extraneous solutions. $$ \sqrt{12-x}=x $$

Step-by-Step Solution

Answer

The solution is \( x = 3 \).

Step 1: Remove the square root by squaring both sides

Begin solving the equation by squaring both sides to eliminate the square root. Start with: \[ \sqrt{12-x} = x \]Square both sides: \[ (\sqrt{12-x})^2 = x^2 \]This simplification gives us: \[ 12-x = x^2 \]

Step 2: Rearrange the equation

Transform the equation into a standard quadratic form. Start with the equation from Step 1:\[ 12 - x = x^2 \]Rearrange the terms to set the equation to zero:\[ x^2 + x - 12 = 0 \]

Step 3: Solve the quadratic equation using factoring

Factor the quadratic equation:\[ x^2 + x - 12 = 0 \]Look for two numbers that multiply to -12 and add to 1. The numbers are 4 and -3. Thus, the factored form is:\[ (x + 4)(x - 3) = 0 \]

Step 4: Find the solutions for x

Set each factor equal to zero to find the possible solutions:1. \( x + 4 = 0 \) gives \( x = -4 \)2. \( x - 3 = 0 \) gives \( x = 3 \)

Step 5: Check for extraneous solutions

Check each potential solution in the original equation \( \sqrt{12-x} = x \):1. For \( x = -4 \): \( \sqrt{12 - (-4)} = \sqrt{16} = 4 eq -4 \) (NOT a solution)2. For \( x = 3 \): \( \sqrt{12 - 3} = \sqrt{9} = 3 = 3 \) (Valid solution)

Key Concepts

Quadratic Equations

Quadratic equations are polynomial equations of degree two. They typically take the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants with \( a eq 0 \). Solving these equations can lead to real or complex solutions, represented graphically by parabolas. To solve them, various methods such as factoring, completing the square, and using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) can be employed. In the context of solving a radical equation like \( \sqrt{12-x} = x \), our goal is to transform the radical equation to a quadratic one by eliminating the square root and rearranging terms.

Factoring

Factoring involves splitting a quadratic equation into simpler binomial expressions, which can be solved individually to find the solutions for \( x \). For the equation \( x^2 + x - 12 = 0 \), we aim to find two numbers that multiply to \(-12\) and add up to \(1\). These numbers are \(4\) and \(-3\). Thus, the factored expression is \((x + 4)(x - 3) = 0\). Factoring is effective when the quadratic is easily reducible, allowing us to identify the roots through straightforward calculations by setting each factor to zero.

Extraneous Solutions

An extraneous solution arises when solving equations with square roots or other functions that are not one-to-one. These solutions do not satisfy the original equation, even though they might appear valid after algebraic manipulations. In our example equation \( \sqrt{12-x} = x \), after solving, we obtain potential solutions \( x = -4 \) and \( x = 3 \). However, substituting \( x = -4 \) back into the original equation does not hold, making \( -4 \) an extraneous solution while \( x = 3 \) remains valid. Checking for extraneous solutions is crucial to ensure the solution set is correct.

Step-by-step Solution

Breaking down complex problems into smaller, manageable steps can enhance understanding and accuracy. The example of \( \sqrt{12-x} = x \) follows a clear step-by-step process:

• First, eliminate the square root by squaring both sides, leading to a quadratic equation.
• Next, rearrange to bring the equation to standard form.
• Then, use factoring to solve the quadratic equation.
• Find potential solutions by setting each factor equal to zero.
• Finally, verify each solution by substituting it back into the original equation to identify extraneous entries.

This systematic approach ensures no step is skipped, making it easier for students to follow and apply to similar problems.