Problem 23 - Elementary and Intermediate Algebra

Question

Solve each equation. Write all proposed solutions. Cross out those that are extraneous. $$ \sqrt{2 x+11}+2=x $$

Step-by-Step Solution

Answer

Solution: \( x = 7 \); -1 is extraneous.

Step 1: Isolate the Square Root

To solve the equation \( \sqrt{2x + 11} + 2 = x \), first isolate the square root term. Subtract 2 from both sides: \( \sqrt{2x + 11} = x - 2 \).

Step 2: Eliminate the Square Root

Square both sides of the equation to remove the square root: \((\sqrt{2x + 11})^2 = (x - 2)^2 \). This simplifies to \( 2x + 11 = x^2 - 4x + 4 \).

Step 3: Rearrange and Simplify

Rearrange the equation to bring all terms to one side: \( 0 = x^2 - 6x - 7 \). This is a quadratic equation.

Step 4: Solve the Quadratic Equation

Solve the quadratic \( x^2 - 6x - 7 = 0 \) using the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Substitute \( a = 1, b = -6, c = -7 \): \( x = \frac{6 \pm \sqrt{(-6)^2 - 4(1)(-7)}}{2(1)} \).

Step 5: Simplify the Quadratic Formula Results

Compute the discriminant: \((-6)^2 - 4 \times 1 \times (-7) = 36 + 28 = 64\). So, \( x = \frac{6 \pm 8}{2} \). This gives two solutions: \( x = \frac{6 + 8}{2} = 7 \) and \( x = \frac{6 - 8}{2} = -1 \).

Step 6: Check for Extraneous Solutions

Substitute \( x = 7 \) back into the original equation to check: \( \sqrt{2(7) + 11} + 2 = 7 \), simplifies to \( \sqrt{25} + 2 = 7 \), which is true: \( 5 + 2 = 7 \). Now check \( x = -1 \): \( \sqrt{2(-1) + 11} + 2 = -1 \), simplifies to \( \sqrt{9} + 2 = -1 \), which is not true: \( 3 + 2 eq -1 \). Mark \( x = -1 \) as extraneous.

Key Concepts

Extraneous Solutions

When solving equations involving square roots, such as \( \sqrt{2x + 11} + 2 = x \), it's common to encounter what's called *extraneous solutions*. These are solutions that appear mathematically valid during the solving process but don't actually satisfy the original equation.

Extraneous solutions typically occur after both sides of the equation have been altered, particularly when squaring both sides to eliminate a square root. This action can introduce new solutions, so it's important to verify each proposed solution by substituting it back into the original equation.

• In our problem, after solving the equation and finding potential solutions \( x = 7 \) and \( x = -1 \), plugging these values back into the original equation shows that \( x = -1 \) doesn't hold true.
• Thus, \( x = -1 \) is deemed extraneous, leaving \( x = 7 \) as the only valid solution.

Always remember: checking your answers against the initial equation is crucial to identify and discard any solutions that don't fit.

Square Root Isolation

Isolating the square root is the first step in solving any equation that contains a square root term. Consider the equation \( \sqrt{2x + 11} + 2 = x \). To solve it, the square root term must stand alone on one side. This ensures that you can easily manage the algebraic operations you need to perform later.

Here's how it's done:

• Subtract 2 from both sides to isolate the square root: \( \sqrt{2x + 11} = x - 2 \). By doing this, you make it easier to apply the next step, which is eliminating the square root.

Once the square root is isolated, you're ready to eliminate it by squaring both sides of the equation. This transformation paves the way for solving the resulting algebraic expressions or equations. Remember that isolating the square root at the outset simplifies the problem-solving process significantly.

Quadratic Formula

The quadratic formula is a powerful tool for solving quadratic equations. After isolating the square root and eliminating it by squaring, we rearranged our equation to \( x^2 - 6x - 7 = 0 \), which is a classic quadratic equation.

Here’s a brief guide on using the quadratic formula:
The quadratic formula is: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). To apply it, first identify the coefficients \( a \), \( b \), and \( c \) from the equation \( ax^2 + bx + c = 0 \).

• In our example, \( a = 1 \), \( b = -6 \), and \( c = -7 \).
• Calculate the discriminant, \( b^2 - 4ac \), which helps determine the nature of the roots.
• Plug these values into the formula to find the values of \( x \).

Our equation solves to \( x = 7 \) and \( x = -1 \). However, this is where we must check for extraneous solutions. Using the quadratic formula efficiently and accurately can help solve any quadratic equation you encounter, provided you verify your solutions.