Problem 24 - Elementary and Intermediate Algebra

Question

Solve each equation. Write all proposed solutions. Cross out those that are extraneous. $$ \sqrt{2 a-3}+3=a $$

Step-by-Step Solution

Answer

The valid solution is \(a = 6\) and \(a = 2\) is extraneous.

Step 1: Isolate the radical expression

Start by isolating the square root on one side of the equation: \( \sqrt{2a-3} = a - 3 \).

Step 2: Eliminate the square root

Square both sides to remove the square root, resulting in: \((\sqrt{2a-3})^2 = (a-3)^2\), which simplifies to \(2a - 3 = a^2 - 6a + 9\).

Step 3: Rearrange into a quadratic equation

Move all terms to one side to form a quadratic equation: \(a^2 - 8a + 12 = 0\).

Step 4: Factor the quadratic equation

Factor the quadratic equation: \(a^2 - 8a + 12 = (a - 2)(a - 6) = 0\).

Step 5: Solve for the potential solutions

Set each factor equal to zero and solve for \(a\): \(a - 2 = 0\) gives \(a = 2\), and \(a - 6 = 0\) gives \(a = 6\). Thus, the proposed solutions are \(a = 2\) and \(a = 6\).

Step 6: Check for extraneous solutions

Substitute each proposed solution back into the original equation to check for extraneous solutions.For \(a = 2\): \(\sqrt{2(2) - 3} + 3 = 2\) simplifies to \(\sqrt{1} + 3 = 2\), which is incorrect, so \(a = 2\) is extraneous.For \(a = 6\): \(\sqrt{2(6) - 3} + 3 = 6\) simplifies to \(\sqrt{9} + 3 = 6\) which is correct, so \(a = 6\) is a valid solution.

Key Concepts

Quadratic Equations

Quadratic equations are fundamental in algebra. They usually appear in the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants. In our example, we have the quadratic equation \( a^2 - 8a + 12 = 0 \). To effectively work through these equations, it is crucial to understand their structure and solve them efficiently.

There are multiple methods to solve quadratic equations:

• Factoring
• Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \)
• Completing the square

Choosing a method depends on the specific equation. In our example, we used factoring because the expression can be easily decomposed into two binomials.

Depending on the discriminant \( b^2 - 4ac \), a quadratic equation can have:

• Two real solutions
• One real solution
• No real solution if it’s negative

Extraneous Solutions

Extraneous solutions are solutions that emerge from the algebraic manipulations of the equation. These solutions are valid given the steps taken, but they do not satisfy the original equation. This often occurs during operations like squaring both sides, which can introduce solutions not initially present.

In the exercise, after solving the quadratic equation \( a^2 - 8a + 12 = 0 \), we got the potential solutions \( a = 2 \) and \( a = 6 \). However, upon substituting back into the original equation \( \sqrt{2a-3}+3=a \), only \( a = 6 \) satisfied it.

To check for extraneous solutions:

• Go back to the original equation
• Substitute each proposed solution
• Verify the equality

This is a pivotal step to ensure the solutions are valid.

Factoring

Factoring is a method used to solve quadratic equations by expressing it as a product of its factors. For the equation \( a^2 - 8a + 12 = 0 \), factoring involves finding two numbers that multiply to 12 and add to -8.

Here's how factoring was done:

• The equation \( a^2 - 8a + 12 \) is split into two binomials: \( (a - 2)(a - 6) = 0 \)
• Each factor is separately set to zero:
• \( a - 2 = 0\) leads to \( a = 2 \)
• \( a - 6 = 0 \) results in \( a = 6 \)

When the factors are correctly identified, solving becomes straightforward by setting each factor equal to zero. This highlights how factoring simplifies solutions and paves the way for further validation against the original equation.