Problem 42 - Algebra 2

Question

Solve. Check for extraneous solutions. \(\sqrt{x}=\sqrt{x-8}+2\)

Step-by-Step Solution

Answer

The solution for the given radical equation \(\sqrt{x}=\sqrt{x-8}+2\) is \(x = 9\).

Step 1: Isolate the radical

First, isolate one of the radical terms. Here, write the equation as \(\sqrt{x} - \sqrt{x-8} = 2\). This leaves a radical on both sides of the equation.

Step 2: Square both sides

Next, square both sides of the equation to eliminate the radical. This yields \((\sqrt{x} - \sqrt{x-8})^2 = 4\), which simplifies further into \(x + x - 8 - 2\sqrt{x(x-8)} = 4\). After simplifying, we get \(2x - 8 - 2\sqrt{x(x-8)} = 4\). Isolating the term containing another radical gives \(2\sqrt{x(x-8)} = 2x - 12\). Dividing by 2, we get \(\sqrt{x(x-8)} = x - 6\).

Step 3: Square again and solve

Square both sides again to remove the radical yielding \(x(x-8) = (x - 6)^2\), which simplifies into \(x^2 - 8x = x^2 - 12x + 36\). Subtract \(x^2 - 12x + 36\) from both sides. This simplifies to \(4x - 36 = 0\). Solve for \(x\) and you get \(x = 9\).

Step 4: Check for extraneous solutions

An extraneous solution is a solution of a transformed equation that is not a solution of the original equation. They are generally introduced by squaring both sides of the equation. Plug \(x = 9\) into the original equation and check if both sides are equal. The right side is \(2\) and the left side is also \(2\). So, \(x = 9\) is a valid solution.

Key Concepts

Extraneous Solutions

Extraneous solutions are solutions that appear in the process of solving an equation, but are not valid for the original equation. These are often introduced in equations involving radicals or fractions. When solving radical equations, this commonly occurs because squaring both sides of an equation can introduce additional solutions that don't satisfy the original radical equation. For instance, in this exercise, after performing operations to solve the radical equation, the solution of the equation appears to be 9. However, it’s crucial to verify that this solution satisfies the original equation to ensure it is not extraneous. To do this, substitute the solution back into the original equation. If both sides of the equation are equal, the solution is valid. In this exercise, substituting 9 into the original equation confirmed that it was a correct solution, as both sides equaled 2. Always remember to check your solutions to avoid accepting extraneous solutions.

Squaring Equations

Squaring equations is an essential step when dealing with radical equations. The primary goal is to eliminate the radicals to transform the equation into a more manageable form, generally a polynomial equation. In this exercise, after isolating the radicals on one side, we squared both sides of the equation

• This turned \( (\sqrt{x} - \sqrt{x-8})^2 = 4 \)
• Which further simplifies to \( 2x - 8 - 2\sqrt{x(x-8)} = 4 \)

By squaring both sides, we aimed to eliminate the radicals to then solve the equation. It's important to proceed carefully with squaring, as it can lead to extraneous solutions, which are not solutions to the original equation but occur in squared versions.

Isolating Radicals

Isolating radicals is the foundation of solving radical equations. Before squaring an equation, one should ensure that the radicals are simplified and isolated on one side of the equation wherever possible. To isolate the radicals effectively:

• Move all terms with radicals to one side of the equation.
• Keep remaining terms without radicals on the opposite side.

In this exercise, we started with \(\sqrt{x} = \sqrt{x-8}+2\). To isolate, we first rewrote it as \(\sqrt{x} - \sqrt{x-8} = 2\),which more clearly separated the radical terms on one side of the equation. Clear isolation is important because it makes it easier to square the equation without introducing more complexity than necessary.