Problem 83 - Intermediate Algebra

Question

Solve each equation. Write all proposed solutions. Cross out those that are extraneous. $$ x=\frac{\sqrt{12 x-5}}{2} $$

Step-by-Step Solution

Answer

Both solutions, \( x = 2.5 \) and \( x = 0.5 \), are valid; none are extraneous.

Step 1: Eliminate the Fraction

Multiply both sides of the equation by 2 to clear the fraction: \[ 2x = \sqrt{12x - 5} \]

Step 2: Eliminate the Square Root

Square both sides of the equation to eliminate the square root: \[ (2x)^2 = (\sqrt{12x - 5})^2 \] This simplifies to: \[ 4x^2 = 12x - 5 \]

Step 3: Rearrange into a Quadratic Equation

Bring all terms to one side to form a quadratic equation: \[ 4x^2 - 12x + 5 = 0 \]

Step 4: Use the Quadratic Formula

Apply the quadratic formula where \( a = 4 \), \( b = -12 \), and \( c = 5 \): \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Substitute these values into the formula: \[ x = \frac{-(-12) \pm \sqrt{(-12)^2 - 4 \cdot 4 \cdot 5}}{2 \cdot 4} \]Simplify the expression under the square root as follows: \[ x = \frac{12 \pm \sqrt{144 - 80}}{8} \]\[ x = \frac{12 \pm \sqrt{64}}{8} \]\[ x = \frac{12 \pm 8}{8} \]

Step 5: Solve for x

There are two possible solutions: \[ x = \frac{12 + 8}{8} = 2.5 \]\[ x = \frac{12 - 8}{8} = 0.5 \]

Step 6: Verify and Identify Extraneous Solutions

Substitute each solution back into the original equation to check for extraneous solutions:1. For \( x = 2.5 \): \[ 2.5 = \frac{\sqrt{12 \cdot 2.5 - 5}}{2} \] \[ 2.5 = \frac{\sqrt{30 - 5}}{2} = \frac{\sqrt{25}}{2} = \frac{5}{2} = 2.5 \] (Valid)2. For \( x = 0.5 \): \[ 0.5 = \frac{\sqrt{12 \cdot 0.5 - 5}}{2} \] \[ 0.5 = \frac{\sqrt{6 - 5}}{2} = \frac{\sqrt{1}}{2} = \frac{1}{2} = 0.5 \] (Valid)

Key Concepts

Quadratic Formula

The quadratic formula is a powerful tool for solving quadratic equations. These are equations of the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants.
The formula is expressed as:

• \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

Understanding how and when to use the quadratic formula is crucial. It allows you to find the values of \(x\) that make the equation true, which are called roots or solutions.
In our example, after manipulating the equation, we reached the form \(4x^2 - 12x + 5 = 0\). Here, \(a = 4\), \(b = -12\), and \(c = 5\). By substituting these values into the quadratic formula, we could solve for \(x\) and find potential solutions.
Similarly, understanding the 'discriminant', \(b^2 - 4ac\), is important. The discriminant determines the nature and number of solutions:

• If positive, there are two real solutions.
• If zero, there is one real solution (a repeated root).
• If negative, there are no real solutions, but two complex solutions.

In this problem, \(b^2 - 4ac\) resulted in 64, indicating two real solutions.

Extraneous Solutions

Extraneous solutions are solutions obtained in the process of solving an equation that do not satisfy the original equation. They often arise when both sides of an equation are squared, or when variables are involved under square roots.
To check for extraneous solutions, always substitute the potential solutions back into the original equation. This will confirm whether they hold true.
In our example, the process of solving involved squaring both sides. This can introduce non-valid solutions. Thus, we had to verify the potential solutions \(x = 2.5\) and \(x = 0.5\).

• For \(x = 2.5\), substitution confirmed it was valid.
• For \(x = 0.5\), substitution also confirmed it was valid.

This step is crucial to ensure the correct final answer without any misleading results.

Algebraic Manipulation

Algebraic manipulation involves rearranging and simplifying equations through operations such as addition, subtraction, multiplication, division, and factorization.
These techniques are essential for isolating terms and simplifying expressions to make equations easier to solve.
At the start of our given problem, we encountered the equation \(x = \frac{\sqrt{12x-5}}{2}\). To eliminate the fraction, we multiplied the entire equation by 2, resulting in \(2x = \sqrt{12x - 5}\).
After that, to remove the square root, both sides were squared, leading us to a standard quadratic equation \(4x^2 - 12x + 5 = 0\).

• This kind of manipulation is necessary to simplify complex equations into forms that are easier to work with.
• It also allowed us to use the quadratic formula effectively.

Being skilled in algebraic manipulation helps in solving not just quadratic equations but also a wide range of mathematical problems.