Problem 25 - Intermediate Algebra

Question

Solve each radical equation. Don't forget, you must check potential solutions. $$ \sqrt{3 x}+6=x $$

Step-by-Step Solution

Answer

The only solution is \( x = 12 \).

Step 1: Isolate the Radical

The equation given is \( \sqrt{3x} + 6 = x \). Start by isolating the square root term on one side of the equation. Subtract 6 from both sides to get: \( \sqrt{3x} = x - 6 \).

Step 2: Eliminate the Square Root

To remove the square root, square both sides of the equation: \( (\sqrt{3x})^2 = (x - 6)^2 \). This simplifies to \( 3x = (x - 6)(x - 6) \).

Step 3: Expand and Simplify

Expand the right side of the equation: \( 3x = x^2 - 12x + 36 \).

Step 4: Rearrange into a Quadratic Equation

Rearrange the equation to form a standard quadratic equation: \( x^2 - 15x + 36 = 0 \).

Step 5: Solve the Quadratic Equation

Factor the quadratic equation: \( (x - 3)(x - 12) = 0 \). This gives the potential solutions \( x = 3 \) and \( x = 12 \).

Step 6: Check Potential Solutions

Substitute back into the original equation to verify: - For \( x = 3 \): \( \sqrt{3 \times 3} + 6 = 3 \), which simplifies to \( 3 + 6 = 3 \). This is incorrect.- For \( x = 12 \): \( \sqrt{3 \times 12} + 6 = 12 \), which simplifies to \( 6 + 6 = 12 \). This checks out.

Key Concepts

Quadratic Equation

In mathematics, a **quadratic equation** is a type of polynomial equation that features a variable raised to the power of two as its highest degree. The general form of a quadratic equation is expressed as:\[ ax^2 + bx + c = 0 \]where \(a\), \(b\), and \(c\) are constants, and \(a eq 0\). Quadratics are essential for solving a variety of problems across algebra and calculus.

• They often arise naturally in situations involving areas, trajectories, and optimization problems.
• While linear equations form straight lines, quadratic equations create parabola shapes when graphed.

To solve a quadratic equation in standard form, you can use several techniques such as factoring, completing the square, or applying the quadratic formula:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \].In the original exercise, factoring was used, which is one of the common methods. The equation \(x^2 - 15x + 36 = 0\) was factored to find the solutions \(x = 3\) and \(x = 12\). This method is particularly useful when the quadratic can be easily expressed as a product of binomials.

Isolating Radicals

The primary step in solving radical equations is **isolating the radical**. It involves rearranging the given equation so that the radical term is alone on one side. This simplifies the process of further solving the equation. For instance, consider the equation:\[ \sqrt{3x} + 6 = x \].The goal is to ignore or "remove" the radicals momentarily. To isolate \(\sqrt{3x}\), you subtract 6 from both sides, obtaining:\[ \sqrt{3x} = x - 6 \].

• This ensures all other constant or variable terms are moved away, making the problem easier to handle.
• Isolating is crucial because it prepares the equation for squaring, which eliminates the radical altogether.
• After isolating, you treat the equation similarly to a regular algebraic equation, setting the stage for the next step.

Having isolated \(\sqrt{3x}\), we can proceed with squaring both sides to eliminate the root, ensuring that we only have polynomial terms left. By managing the equation in such a structured way, you lay the groundwork for easily completing subsequent operations.

Checking Solutions

Once you've solved a radical equation, **checking solutions** is a vital step to verify their correctness. Solving operations might sometimes introduce extraneous solutions—false results that don't satisfy the original conditions of the equation. Let's see this applied:

• After finding potential solutions from the quadratic \((x - 3)(x - 12) = 0\), the values of \(x\) obtained were 3 and 12.
• To check, substitute each back into the original equation \(\sqrt{3x} + 6 = x\).

For \(x = 3\), substituting gives:\[ \sqrt{3 \times 3} + 6 = 3 \]simplifying to:\\[ 3 + 6 = 3 \]which is incorrect. Thus, \(x = 3\) is not a valid solution.
For \(x = 12\), substituting provides:\[ \sqrt{3 \times 12} + 6 = 12 \]simplifying to:\[ 6 + 6 = 12 \], confirming it checks out.

• Only solutions that satisfy the original equation, post-substitution, should be accepted.
• Checking reinforces understanding and ensures accuracy, preventing acceptance of misleading results.

This practice ensures confidence in your final answers and underscores the importance of revisiting your solutions with the original context in mind.