Problem 24 - Precalculus

Question

Solve the equation or inequality. $$\sqrt{x-2}+\sqrt{x-5}=3$$

Step-by-Step Solution

Answer

The solution to the equation is \( x = 6 \).

Step 1: Isolate one radical expression

To solve the equation \( \sqrt{x-2} + \sqrt{x-5} = 3 \), let's first isolate one of the square root terms. In this case, isolate \( \sqrt{x-2} \) by moving \( \sqrt{x-5} \) to the other side: \( \sqrt{x-2} = 3 - \sqrt{x-5} \).

Step 2: Square both sides of the equation

To eliminate the square root on the left side, square both sides of the equation: \( (\sqrt{x-2})^2 = (3 - \sqrt{x-5})^2 \). This simplifies to \( x - 2 = 9 - 6\sqrt{x-5} + (x-5) \).

Step 3: Simplify the equation

Combine like terms on the right side: \( x - 2 = x + 4 - 6\sqrt{x-5} \). Subtract \( x \) from both sides to get \( -2 = 4 - 6\sqrt{x-5} \).

Step 4: Isolate the remaining radical expression

Move all terms except the radical to the left side: \( -2 - 4 = -6\sqrt{x-5} \), which simplifies to \( -6 = -6\sqrt{x-5} \).

Step 5: Solve for the square root

Divide both sides by -6 to isolate the square root: \( 1 = \sqrt{x-5} \).

Step 6: Square both sides again

Square both sides to remove the square root: \( 1^2 = (\sqrt{x-5})^2 \), which results in \( 1 = x - 5 \).

Step 7: Solve for \( x \)

Add 5 to both sides of the equation: \( 1 + 5 = x \). So, \( x = 6 \).

Step 8: Check the solution

Substitute \( x = 6 \) back into the original equation to check: \( \sqrt{6-2} + \sqrt{6-5} = 3 \) simplifies to \( \sqrt{4} + \sqrt{1} = 3 \), which is \( 2 + 1 = 3 \). This satisfies the original equation.

Key Concepts

Square Root

The square root is a mathematical operation that, when applied to a number, returns the value that, when multiplied by itself, gives the original number. In other words, if you have a number like 4, the square root is 2, because 2 multiplied by 2 equals 4. Square roots are usually represented by the radical symbol (√). Understanding square roots is essential because they are a common part of many equations, especially quadratic and radical equations.

In the context of solving radical equations, square roots are often presented in a form such as \( \sqrt{x-2} \), which means you're looking for a number that, when squared, gives \( x-2 \). This makes them crucial when you're solving equations that involve radicals.

Isolating Radicals

Isolating radicals is a crucial step in solving radical equations. The goal is to have one radical expression alone on one side of the equation. This makes it easier to eliminate the radical terms by squaring both sides later. For example, in an equation like \( \sqrt{x-2} + \sqrt{x-5} = 3 \), you might first move one square root term to the opposite side.

This gives us \( \sqrt{x-2} = 3 - \sqrt{x-5} \). Having isolated the \( \sqrt{x-2} \), it becomes simpler to handle, especially once it comes time to square each side of the equation. Isolating radicals helps in simplifying complex expressions, making the subsequent solving process manageable.

Solving Equations

Solving equations that involve square roots or other radicals requires a robust understanding of algebraic techniques. After isolating a radical and simplifying the equation whenever possible, the next step typically involves squaring both sides to eliminate the radical. For instance, from \( \sqrt{x-2} = 3 - \sqrt{x-5} \), we square both sides to remove the square root:

\((\sqrt{x-2})^2 = (3-\sqrt{x-5})^2.\)

This simplifies to \( x-2 = 9 - 6\sqrt{x-5} + (x-5) \). By working through these algebraic steps, we systematically break down the equation until we find a more straightforward form, such as a linear expression or basic arithmetic equation.

Each solution step should be done cautiously, as any miscalculation can lead to incorrect outcomes. Solving these equations is systematic, involving careful steps of simplification, isolation, and operation reversal.

Checking Solutions

After finding a potential solution for an equation, it is crucial to verify it within the original equation. Checking the solution ensures that it satisfies all parts of the original problem, which is especially important for equations involving radicals. Roots sometimes bring extraneous solutions; values that emerge during solving but do not satisfy the original equation.

For the equation \( \sqrt{x-2} + \sqrt{x-5} = 3 \), after finding \( x = 6 \), plugging this back verifies correctness: \( \sqrt{6-2} + \sqrt{6-5} = 3 \) simplifies to \( \sqrt{4} + \sqrt{1} = 3 \), confirming \( 2 + 1 = 3 \).

Verification is a vital step to ensure the integrity and accuracy of solutions in mathematics, particularly with the risk of extraneous solutions when dealing with radicals.