Chapter 6: Problem 25
Solve each radical equation. Don't forget, you must check potential solutions. $$ \sqrt{3 x}+6=x $$
Short Answer
Expert verified
The only solution is \( x = 12 \).
Step by step solution
01
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: \( \sqrt{3x} = x - 6 \).
02
Eliminate the Square Root
To remove the square root, square both sides of the equation: \( (\sqrt{3x})^2 = (x - 6)^2 \). This simplifies to \( 3x = (x - 6)(x - 6) \).
03
Expand and Simplify
Expand the right side of the equation: \( 3x = x^2 - 12x + 36 \).
04
Rearrange into a Quadratic Equation
Rearrange the equation to form a standard quadratic equation: \( x^2 - 15x + 36 = 0 \).
05
Solve the Quadratic Equation
Factor the quadratic equation: \( (x - 3)(x - 12) = 0 \). This gives the potential solutions \( x = 3 \) and \( x = 12 \).
06
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.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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.
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.
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:
For \(x = 12\), substituting provides:\[ \sqrt{3 \times 12} + 6 = 12 \]simplifying to:\[ 6 + 6 = 12 \], confirming it checks out.
- 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 \(x = 12\), 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.