Problem 25 - College Algebra with Corequisite Support

Question

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

Step-by-Step Solution

Answer

The solution is \( x = 3 \).

Step 1: Isolate the Square Root

We start with the equation given: \( \sqrt{12-x} = x \). The square root is already isolated on one side of the equation, so we can proceed to the next step of eliminating the square root by squaring both sides.

Step 2: Square Both Sides

Square both sides of the equation to eliminate the square root:\( (\sqrt{12-x})^2 = x^2 \). This simplifies to:\( 12-x = x^2 \).

Step 3: Rearrange the Equation

Rearrange the equation to set it to zero:\( x^2 + x - 12 = 0 \). This is a quadratic equation.

Step 4: Solve the Quadratic Equation

Factor the quadratic equation:\( x^2 + x - 12 = (x - 3)(x + 4) = 0 \). This gives the solutions:\( x - 3 = 0 \) or \( x + 4 = 0 \). So, \( x = 3 \) or \( x = -4 \).

Step 5: Check the Solutions

Substitute each solution back into the original equation to verify if they hold true:- For \( x = 3 \): \( \sqrt{12 - 3} = 3 \rightarrow \sqrt{9} = 3 \rightarrow 3 = 3 \), which is true.- For \( x = -4 \): \( \sqrt{12 - (-4)} = -4 \rightarrow \sqrt{16} = -4 \rightarrow 4 eq -4 \), which is false.Thus, the only valid solution is \( x = 3 \).

Key Concepts

Quadratic Equations

Quadratic equations are a fundamental concept in algebra and appear often when working with radical equations. These equations have a standard form: \[ ax^2 + bx + c = 0 \]where \(a\), \(b\), and \(c\) are constants, and \(x\) represents the variable. In the given exercise, after eliminating the square root, the equation \(12-x = x^2\) was rearranged to the quadratic form \(x^2 + x - 12 = 0\). Quadratic equations can be solved using various methods such as:

• Factoring: Finding two binomials that generate the quadratic equation when multiplied.
• Quadratic Formula: Using the formula \(x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\) to find the solutions for \(x\).
• Completing the Square: Rewriting the equation to form a perfect square trinomial.

In our exercise, factoring was used: \((x - 3)(x + 4) = 0\). This breaks it into two simpler equations: \(x - 3 = 0\) and \(x + 4 = 0\), leading to possible solutions \(x = 3\) and \(x = -4\). It's essential to test both solutions due to the nature of radical equations to avoid incorporating extraneous solutions.

Extraneous Solutions

Sometimes, during the algebraic manipulation of radical equations, solutions called "extraneous solutions" may arise. These are results that fit the transformed version of an equation but do not satisfy the original equation. They often occur when both sides of an equation are squared, a step which can potentially introduce false solutions.

In our exercise, we initially squared the equation to simplify \((\sqrt{12-x})^2 = x^2\) into \(12-x = x^2\). After solving, substitute back the possible solutions into the original radical equation \(\sqrt{12-x} = x\) to confirm their validity.
For instance:

• For \(x = 3\), substituting back shows \(\sqrt{12-3} = 3\), indeed valid.
• For \(x = -4\), it results in \(\sqrt{12-(-4)} = -4\), which simplifies to \(\sqrt{16} = -4\), an impossibility, indicating \(x = -4\) is extraneous.

Neither positive nor negative square roots can equate to a negative number, thus only \(x = 3\) is a valid solution here. Such checks are crucial to ensure no extraneous solutions finalize as accepted answers.

Algebraic Manipulation

Algebraic manipulation is a powerful toolkit in solving equations. It involves carefully applying algebraic operations to maintain equality while reshaping the equation to find solutions easily. For solving radical equations, several strategies help manage the radical symbol:

• Isolating the Radical: Begin by getting the square root alone on one side of the equation. This exercise already began with \(\sqrt{12-x} = x\), making it ready for squaring.
• Squaring Both Sides: This removes the square root but introduces the potential for extraneous solutions, as shown in this exercise.
• Rearranging: Align terms to form convenient expressions, such as quadratics, that are easier to solve. We converted \(12-x = x^2\) into \(x^2 + x - 12 = 0\).

Rearranging stages equations into standard forms, such as linear or quadratic, where known solving techniques like factoring or using the quadratic formula can be effectively applied. Each manipulation step has the universal aim of either simplifying the expression or achieving a solvable format, all the while cross-verifying that transformations preserve original solution constraints. Hence, strategic manipulation not only simplifies the process but enhances precision when solving radical equations.