Problem 13 - College Algebra Essentials

Question

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

Step-by-Step Solution

Answer

The solution to the equation \(\sqrt{x+3}=x-3\) is \( x = 6 \).

Step 1: Square Both sides

Square both sides of the equation to get rid of the square root: \[ (\sqrt{x+3})^2 = (x-3)^2 \] This simplifies to: \[ x+3 = x^2 - 6x + 9 \]

Step 2: Reorganize the equation

The equation obtained from Step 1 is a quadratic equation. Reorganize it into the standard form of a quadratic equation by moving all terms to one side: \[ 0 = x^2 - 7x + 6 \]

Step 3: Solve the quadratic equation

Now, factor the quadratic equation to find the possible values for x: \[ 0 = (x-6)(x-1) \] From this, the possible solutions are \( x = 6 \) or \( x = 1 \)

Step 4: Check for extraneous solutions

Check these solutions by substituting them back into the original equation. For \( x = 6 \): \[ \sqrt{6 + 3} = 6 - 3 \] \[ 3 = 3 \] So, \( x = 6 \) is a solution. For \( x = 1 \): \[ \sqrt{1 + 3} != 1 - 3 \] \[ 2 != -2 \] So, \( x = 1 \) is an extraneous solution and is not a solution to the original equation.

Key Concepts

Factoring Quadratics

Factoring quadratics is a powerful technique used in solving quadratic equations. Quadratics are equations of the form \( ax^2 + bx + c = 0 \), and factoring involves writing the quadratic as a product of its factors. This process simplifies solving the equation.
Let's consider the equation from the exercise: \( x^2 - 7x + 6 = 0 \). To factor this quadratic, we look for two numbers that multiply to the constant term (6 in this case) and add up to the linear coefficient (-7).

• Identify the constant term, which is 6.
• Find two numbers that multiply to 6 and add to -7. These numbers are -6 and -1.

Therefore, we can rewrite the equation as \( (x - 6)(x - 1) = 0 \).
The solutions to the quadratic are the values of \( x \) that make each factor equal to zero. Solving \( (x - 6) = 0 \) gives \( x = 6 \), and solving \( (x - 1) = 0 \) gives \( x = 1 \). These two values are potential solutions.

Checking Extraneous Solutions

When solving equations, especially those involving square roots or other operations, we may encounter extraneous solutions. Extraneous solutions arise when an operation, such as squaring, modifies the original equation. It's imperative to substitute these solutions back into the initial equation to verify their validity.
In the given exercise, we derived potential solutions \( x = 6 \) and \( x = 1 \) after manipulating the equation. Let's check each:

• For \( x = 6 \), substitute back into the original equation: \( \sqrt{6 + 3} = 6 - 3 \). Simplifying both sides gives \( 3 = 3 \), confirming \( x = 6 \) is valid.
• For \( x = 1 \), substitute back as well: \( \sqrt{1 + 3} = 1 - 3 \). This yields \( 2 = -2 \), clearly not true, marking \( x = 1 \) as extraneous.

Checking solutions prevents incorrect answers from spoiling the problem-solving process. Always confirm solutions to ensure they fit the original problem.

Square Root Equations

Square root equations commonly involve an equation with a variable inside a square root. To solve these, the first step is to isolate the square root on one side, then eliminate it by squaring both sides of the equation.
In the original exercise, we had \( \sqrt{x+3} = x-3 \). To solve this:

• Square both sides to remove the square root: \( (\sqrt{x+3})^2 = (x-3)^2 \).
• This yields \( x + 3 = x^2 - 6x + 9 \), a quadratic equation.

Squaring each side can potentially introduce non-valid solutions like those seen in extraneous solutions.
After eliminating the square root, proceed with solving the resultant quadratic equation. Always remember to verify your solutions against the original equation to rule out any errors.