Problem 12 - Algebra 1: Concepts and Skills

Question

Solve the equation. Check for extraneous solutions. $$ x=\sqrt{x+12} $$

Step-by-Step Solution

Answer

The solution to the equation \(x= \sqrt{x+12}\) is \(x = 4\). The potential solution \(x = -3\) is extraneous and does not satisfy the original equation.

Step 1: Squaring both sides

To get rid of the square root on the right side, square both sides of the equation. This gives: \(x^2 = x + 12\)

Step 2: Rearrange into a Quadratic Equation

Rearrange the equation into a standard quadratic form \(ax^2 + bx + c = 0\). This gives \(x^2 - x - 12 = 0\)

Step 3: Factor the Quadratic Equation

Factor the quadratic equation. The factors of -12 that sum to -1 are -4 and 3. So, the factored form of the equation is \((x-4)(x+3) = 0\)

Step 4: Solve for x

Set each factor equal to zero and solve for x. This gives two solutions: \(x-4 = 0\) gives \(x = 4\) and \(x + 3 = 0\) gives \(x = -3\)

Step 5: Check for Extraneous Solutions

Check these solutions in the original equation to make sure they are not extraneous. Substituting \(x = 4\) into the original equation gives \(4 = \sqrt{4+12}\), which simplifies to \(4 = \sqrt{16}\), which is true. Substituting \(x = -3\) into the original equation gives \(-3 = \sqrt{-3+12}\), which simplifies to \(-3 = \sqrt{9}\), which is not true. Therefore, \(x = -3\) is an extraneous solution and is discarded.

Key Concepts

Understanding Quadratic Equations

A quadratic equation is a type of polynomial equation that takes the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants. Quadratics are fundamental in algebra and involve terms up to the second degree. The solutions to these equations correspond to the values of \(x\) that make the equation true.

Generally, there are several methods to solve quadratic equations, including:

• Factoring: Expressing the equation as a product of its factors.
• Completing the square: Manipulating the equation to make it a perfect square trinomial.
• Quadratic formula: Using the formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) to find the roots.

In the exercise, the equation was rearranged and then factored to find the solutions. The nature of quadratics often allows for two solutions, reflecting the idea that a parabola (graph of a quadratic equation) can intersect the x-axis at up to two points.

Dealing with Extraneous Solutions

Extraneous solutions are results that may arise in the process of solving equations, particularly when dealing with square roots, rational expressions, or logarithms. Just because a solution solves a transformed equation doesn't mean it solves the original equation.

In our exercise, after solving \(x^2 = x + 12\) by factoring, we found potential solutions \(x = 4\) and \(x = -3\). However, upon substituting them back into the original equation \(x = \sqrt{x+12}\), only \(x = 4\) satisfied the equation, while \(x = -3\) did not. This substitution step is crucial to identify and discard extraneous solutions.

Checking for extraneous solutions is important to ensure the solutions are valid in the context of the original problem. Ignoring this step can lead to incorrect conclusions.

Factoring Polynomials Effectively

Factoring polynomials is a powerful technique often used in solving quadratic equations. It involves expressing a polynomial as a product of simpler polynomials. For a quadratic equation like \(x^2 - x - 12 = 0\), the goal is to find two numbers that multiply to the constant term \(-12\) and add up to the linear coefficient \(-1\).

Steps to factor a quadratic equation include:

• Identify the constant term and the linear coefficient from the quadratic equation.
• Find two numbers that produce the constant term when multiplied and add up to the linear coefficient.
• Rewrite the quadratic equation in its factored form using these numbers.

In the exercise, the numbers \(-4\) and \(3\) worked because \((-4) \times 3 = -12\) and \(-4 + 3 = -1\). Therefore, \((x-4)(x+3) = 0\) is the factored form, from which the solutions \(x = 4\) and \(x = -3\) are derived by setting each factor to zero. Factoring is a skill that simplifies solving many equations, emphasizing the importance of recognizing pattern structures in polynomials.