Problem 73 - Intermediate Algebra

Question

Solve each equation. Write all proposed solutions. Cross out those that are extraneous. $$ \sqrt{x+5}+\sqrt{x-3}=4 $$

Step-by-Step Solution

Answer

The solution is \( x = 4 \); it is valid without extraneous solutions.

Step 1: Isolate one of the square roots

To solve the equation \( \sqrt{x+5} + \sqrt{x-3} = 4 \), start by isolating one of the square root terms. Let's isolate \( \sqrt{x+5} \) by first rewriting the equation as follows: \( \sqrt{x+5} = 4 - \sqrt{x-3} \).

Step 2: Square both sides to eliminate the square root

Next, square both sides of the equation to eliminate the square root on the left. This will give us: \( (\sqrt{x+5})^2 = (4 - \sqrt{x-3})^2 \). Simplify to get: \( x + 5 = 16 - 8\sqrt{x-3} + (x-3) \).

Step 3: Simplify and rearrange the equation

Simplify the equation further by combining like terms: \( x + 5 = x + 13 - 8\sqrt{x-3} \). To further simplify, subtract \( x + 5 \) from both sides, resulting in \( 0 = 8 - 8\sqrt{x-3} \).

Step 4: Isolate and solve for the remaining square root

Isolate \( \sqrt{x-3} \) in the equation obtained: \( 8\sqrt{x-3} = 8 \). Divide both sides by 8 to get \( \sqrt{x-3} = 1 \).

Step 5: Square both sides again

Square both sides of the equation \( \sqrt{x-3} = 1 \) to eliminate the square root, obtaining \( x-3 = 1 \).

Step 6: Solve for \( x \)

Add 3 to both sides of the equation to solve for \( x \), yielding \( x = 4 \).

Step 7: Verify proposed solution and identify extraneous solutions

Substitute \( x = 4 \) back into the original equation to verify: \( \sqrt{4+5} + \sqrt{4-3} = 3 + 1 = 4 \), which is true. Thus, \( x = 4 \) is a valid solution. There are no extraneous solutions in this case.

Key Concepts

Square Root Isolation

When faced with an equation containing square roots, the first step usually involves isolating one of the square root terms. This is important because it prepares the equation for the next steps in solving – namely, squaring the equation to remove the square roots. By isolating one square root on one side of the equation, we simplify the equation considerably.
In the exercise provided, we began with the equation \( \sqrt{x+5} + \sqrt{x-3} = 4 \). The goal was to isolate either \( \sqrt{x+5} \) or \( \sqrt{x-3} \). Isolation turns a more complex structure into something more manageable:

• Subtract \( \sqrt{x-3} \) from both sides to get \( \sqrt{x+5} = 4 - \sqrt{x-3} \).

By doing this, we focus on one of the roots at a time, which cuts down the complexity and prepares us to square both sides of the equation in the subsequent steps.

Extraneous Solutions

An important consideration when solving radical equations is the possibility of extraneous solutions. Extraneous solutions are results that emerge naturally from the manipulation of the equation but do not actually satisfy the original equation.
This can occur when squaring both sides of an equation as it can introduce new solutions. Since squaring is a non-reversible operation in the context of negative and positive roots, some solutions that may not work in the original equation appear valid in the transformed versions. To handle this:

• Always substitute each solution back into the original equation to check for validity.

In our exercise, after isolating and squaring appropriately, we found \( x = 4 \). Upon substitution back into \( \sqrt{x+5} + \sqrt{x-3} = 4 \), it was confirmed that \( x = 4 \) is indeed a correct solution, meaning we encountered no extraneous solutions this time.

Verification of Solutions

Verification is a critical final step in solving radical equations. As with any attempt to solve an equation where transformative steps like squaring are involved, verifying ensures that the solutions derived genuinely satisfy the original condition.
Verification involves a direct substitution of the derived solution back into the original equation:

• In our case, substituting \( x = 4 \) back into \( \sqrt{x+5} + \sqrt{x-3} \) confirms the equality: \( \sqrt{9} + \sqrt{1} = 3 + 1 = 4 \).
• Since the simplified left-hand side equals 4, which was the original right-hand side, \( x = 4 \) is confirmed.

Through verification, any doubts about the correctness and authenticity of solutions are cleared up. This final step distinguishes valid solutions from those that were potentially introduced through the manipulation of the equation – ensuring complete confidence in the answers derived.