Problem 74 - Elementary and Intermediate Algebra

Question

Solve each equation. Write all proposed solutions. Cross out those that are extraneous. $$ \sqrt{b+7}-\sqrt{b-5}=2 $$

Step-by-Step Solution

Answer

The solution is \( b = 9 \) and it is not extraneous.

Step 1: Isolate One Square Root Term

Start by isolating one of the square root terms on one side of the equation. For the equation \( \sqrt{b+7} - \sqrt{b-5} = 2 \), add \( \sqrt{b-5} \) to both sides to get: \[ \sqrt{b+7} = \sqrt{b-5} + 2 \]

Step 2: Square Both Sides

Square both sides to eliminate the square root on the left side of the equation: \[ (\sqrt{b+7})^2 = (\sqrt{b-5} + 2)^2 \] This simplifies to: \[ b + 7 = b - 5 + 4\sqrt{b-5} + 4 \]

Step 3: Simplify the Equation

Simplify the equation from the previous step: \[ b + 7 = b - 1 + 4\sqrt{b-5} \] Subtract \( b - 1 \) from both sides to isolate the term with the square root: \[ 8 = 4\sqrt{b-5} \]

Step 4: Solve for the Second Square Root

Divide both sides by 4 to solve for the square root: \[ 2 = \sqrt{b-5} \] Now square both sides again to eliminate the square root: \[ 4 = b - 5 \]

Step 5: Solve for b

Add 5 to both sides of the equation to solve for \( b \): \[ b = 9 \]

Step 6: Check for Extraneous Solutions

Substitute \( b = 9 \) back into the original equation to ensure it doesn't lead to an invalid solution: \[ \sqrt{9+7} - \sqrt{9-5} = 2 \] \[ \sqrt{16} - \sqrt{4} = 4 - 2 = 2 \] Since both sides of the equation equal each other, \( b = 9 \) is a valid solution.

Key Concepts

Extraneous Solutions

When solving radical equations, it is important to watch out for extraneous solutions. These are solutions that appear valid during the solving process but do not satisfy the original equation. They often arise when both sides of an equation are squared. Here's why they can happen:- Squaring both sides of an equation can introduce new solutions that weren't present in the original.- Certain operations, like squaring, aren't reversible in a straightforward manner without potentially altering the solution set.To identify extraneous solutions, always substitute proposed answers back into the original equation. If substituting a solution results in an equality (both sides of the equation are the same), then the solution is valid. If not, it is extraneous and should be discarded. For example, after determining that \( b = 9 \), we substitute it back into the original equation \( \sqrt{b+7} - \sqrt{b-5} = 2 \). The equation holds true, so \( b = 9 \) is not extraneous.

Square Root Isolation

Square root isolation is a key step when solving equations involving radicals. It helps to simplify the equation and reduce it to a form that is easier to handle. The first goal is to have one square root on one side of the equation. Consider the equation \( \sqrt{b+7} - \sqrt{b-5} = 2 \). By adding \( \sqrt{b-5} \) to both sides, we isolate the first radical: \[ \sqrt{b+7} = \sqrt{b-5} + 2 \]The isolated square root can now be managed more easily, allowing you to perform further operations like squaring both sides.Steps to isolate a square root:

• Move all terms, except one with a square root, to the opposite side.
• Ensure the isolated radical is alone on one side of the equation.

This isolation is crucial before any squaring because it minimizes errors and makes the subsequent steps simpler.

Squaring Both Sides

Once a square root is isolated, you can eliminate it by squaring both sides of the equation. Squaring is a handy technique, but it must be done carefully to avoid mistakes. Step-by-step process:

• Square both sides of the isolated equation, ensuring each side is squared independently.
• For example, from \( \sqrt{b+7} = \sqrt{b-5} + 2 \), squaring both sides gives: \[ (\sqrt{b+7})^2 = (\sqrt{b-5} + 2)^2 \]
• Calculate each side. The left side becomes \( b + 7 \). The right side expands to \( b - 5 + 4\sqrt{b-5} + 4 \).

This action results in the elimination of the square root on the left, simplifying the equation significantly. However, be cautioned— squaring can introduce extraneous solutions, so always verify by checking any solutions against the original equation after solving.