Problem 30 - Algebra 2

Question

Solve. Check for extraneous solutions. \(1=(3+x)^{\frac{1}{2}}\)

Step-by-Step Solution

Answer

The solution for \(x\) is -2, which is not an extraneous solution, as it is valid in the original equation.

Step 1: Remove the square root

First, in order to remove the square root, square both sides of the equation. So, \((1)^{2}=((3+x)^{\frac{1}{2}})^{2}\), simplifying gives \(1=3+x\).

Step 2: Solve for x

Next, to solve for \(x\), subtract 3 from both sides of the equation. This gives \(x=1-3=-2\).

Step 3: Check for extraneous solutions

Lastly, substitute \(x=-2\) into the original equation \(1=(3+x)^{\frac{1}{2}}\). That yields \(1=(3+-2)^{\frac{1}{2}}=1\). Since the left side is equal to the right side, \(x=-2\) is a valid solution, not an extraneous one.

Key Concepts

Extraneous Solutions

When solving equations involving radicals, it's crucial to check for extraneous solutions. An extraneous solution is a solution derived from the algebraic manipulation of the equation, which doesn’t satisfy the original equation. Why do these occur? Often when both sides of an equation are squared, new solutions that didn't exist before the operation can appear. Without proper verification, these incorrect solutions could be mistakenly accepted as true.

• Squaring both sides, especially with radical equations, can introduce these errors.
• Ensuring solutions are valid requires substituting them back into the original equation.
• Eliminate solutions that don't satisfy the original equation as they are extraneous.

For example, in the exercise, after solving for \( x = -2 \), we rechecked by plugging \( x = -2 \) back into the original equation \( 1 = (3 + x)^{\frac{1}{2}} \), confirming it was a valid solution.

Radical Equations

Radical equations are equations where the variable is under a radical symbol, like a square root or cube root. Solving these involves isolating the radical and eliminating it by raising both sides of the equation to a power, equal to the radical's index. This results in a polynomial equation that's usually easier to handle.

• Identify the radical part of the equation, and decide the power to raise the equation to eliminate the radical.
• Ensure all terms with the variable are on one side before performing the operation.
• Radical equations often require checking for extraneous solutions due to the manipulation involved.

In our example, the square root was eliminated by squaring both sides, leading us to the equation \( 1 = 3 + x \), which was much simpler to solve.

Algebraic Manipulation

Algebraic manipulation is a method of rearranging and simplifying equations to isolate the variable of interest. This can involve a variety of operations including addition, subtraction, multiplication, division, and use of exponents. The goal is to perform operations that gradually make the equation simpler and isolate the variable.

• Start with simplifying both sides of the equation, removing radicals or fractions if necessary.
• Use inverse operations strategically to cancel terms, thus isolating the variable.
• After solving the equation, always reassess to ensure no errors occurred during manipulation.

In the provided exercise, after removing the radical, we subtracted 3 from both sides of the equation. This simple manipulation isolated \( x \), giving us \( x = -2 \). Such manipulations are fundamental in moving towards the solution.