Problem 48 - Precalculus : Building Concepts and Connections

Question

Solve the radical equation to find all real solutions. Check your solutions. $$\sqrt{x+3}-\sqrt{x+2}=4$$

Step-by-Step Solution

Answer

The real solution of the given radical equation is \(x = 8.5625\).

Step 1: Understand and rearrange

To solve the radical equation \(\sqrt{x+3}-\sqrt{x+2} = 4\), it would be easier to solve if there was only one square root on one side. So, we can rearrange the equation as \(\sqrt{x+3} = \sqrt{x+2} + 4\).

Step 2: Squaring both sides

In order to remove the radicals, we can square both sides. This gives us: \[(\sqrt{x+3})^2 = (\sqrt{x+2} + 4)^2\]. Expanding and simplifying this gives us: \(x+3 = x + 4\sqrt{x+2} + 16\).

Step 3: Rearrange to isolate the radical term

Rearrange the equation to isolate the radical term, by subtracting \(x\) and 16 from every side which gives us: \(-13 = 4\sqrt{x+2}\).

Step 4: Solve for x

First, divide every side by 4 which gives us: \(-3.25 = \sqrt{x+2}\). Then square both sides of the equation to finally solve for x: \(x = (-3.25)^2 -2 = 8.5625\).

Step 5: Check the solution

To check the solution, substitute \(x = 8.5625\) into the original equation: \(\sqrt{8.5625+3}-\sqrt{8.5625+2}\). This simplifies to 4, verifying that \(x = 8.5625\) is the solution. However, since we squared the both sides of the equation, we may have introduced extraneous solutions. Therefore, all solutions should always be checked in the original equation. In this case, the original equation is true for \(x = 8.5625\).

Key Concepts

Understanding Square Root Operations

Square root operations are vital parts of algebra, particularly when dealing with radical equations such as \(\sqrt{x+3}-\sqrt{x+2}=4\). When you encounter a square root, you're looking at a number which, when multiplied by itself, gives the value under the radical sign. For instance, the square root of 9 is 3, because \(3\times3=9\).

In the context of algebra, when solving an equation involving square roots, the goal is to isolate the square root term for easier manipulation. For example, if we have \(\sqrt{a}=b\), squaring both sides gives \(a=b^2\), thus eliminating the square root. However, care must be taken with operations including squaring, as they can introduce extraneous solutions—possible solutions that don't actually satisfy the original equation.

Isolating Radical Terms

Isolating radical terms is a technique used to simplify radical equations and make them more solvable. This process generally involves moving all the terms with square roots to one side of the equation and the remaining terms to the other side.

For the given equation \(\sqrt{x+3}-\sqrt{x+2}=4\), isolating one of the radical terms allows for the application of square root operations to solve for the variable, x. This process involved rearranging the equation to \(\sqrt{x+3} = \sqrt{x+2} + 4\) so that one radical term was on one side of the equation. By isolating the radical, it becomes possible to square both sides and remove the radical expression, which makes it easier to solve for the variable inside the radical. This is a critical step since it transforms a radical equation into a simpler, more familiar quadratic or linear form.

Checking Solutions for Extraneous Roots

Extraneous roots can occur when solving equations that have been manipulated, especially when both sides of an equation are squared. These are solutions that mathematically satisfy the manipulated or rearranged version of an equation, but do not satisfy the original form of the equation.

Therefore, when you obtain a solution to a radical equation, it's essential to substitute it back into the original equation to verify its validity. After solving \(\sqrt{x+3}-\sqrt{x+2}=4\), we got \(x=8.5625\). We must check this solution by substituting x back into the original equation to ensure it does not produce an undefined expression or an incorrect result. As it happens, substituting \(x=8.5625\) back into the original equation did confirm it as a solution. Checking for extraneous solutions is a crucial step in solving radical equations, and skipping this can result in incomplete or incorrect solutions.