Problem 16 - Algebra 2 and Trigonometry

Question

In \(3-38,\) solve each equation for the variable, check, and write the solution set. $$ 8+\sqrt{2 x-1}=15 $$

Step-by-Step Solution

Answer

The solution set is \(\{25\}\).

Step 1: Isolate the Square Root

Start by isolating the square root on one side of the equation. Subtract 8 from both sides of the equation: \[\sqrt{2x-1} = 15 - 8\] This simplifies to: \[\sqrt{2x-1} = 7\]

Step 2: Square Both Sides

To eliminate the square root, square both sides of the equation. This step involves raising both sides to the power of 2: \[(\sqrt{2x-1})^2 = 7^2\] This simplifies to: \[2x - 1 = 49\]

Step 3: Solve for x

Now, solve for \(x\) by adding 1 to both sides and then dividing by 2: \[2x = 49 + 1\] \[2x = 50\] Divide both sides by 2: \[x = \frac{50}{2}\] \[x = 25\]

Step 4: Check the Solution

Substitute \(x = 25\) back into the original equation to check for validity: Original Equation:\[8 + \sqrt{2(25) - 1} = 15\] Calculate inside the square root first:\[8 + \sqrt{50 - 1} = 15\] \[8 + \sqrt{49} = 15\] Simplify:\[8 + 7 = 15\] This confirms that the solution \(x = 25\) satisfies the equation.

Step 5: Write the Solution Set

The solution set of the equation is all values of \(x\) that satisfy it. Since we have verified that \(x = 25\) satisfies the equation, the solution set is: \[\{25\}\]

Key Concepts

Solution Set

When we solve an equation, we are finding a value or set of values that make the equation true. This collection of values is known as the **solution set**. In our exercise, we discovered that the solution for the equation is when \(x = 25\). This single value makes the original equation \(8 + \sqrt{2x-1} = 15\) true, so our solution set is given as \(\{25\}\). It's important to always express your final answer as a solution set, especially when dealing with equations, to clearly communicate the results.

Checking Solutions

Verifying our solutions is a crucial part of solving equations. After finding \(x = 25\), it's wise to substitute it back into the original equation to ensure that it satisfies all parts of the equation.
By doing this, we confirmed that:

• The substituted value results in both sides of the equation being equal.
• It eliminates any chance of a mathematical error during our calculations.

In this case, when substituting \(x = 25\) back into \(8 + \sqrt{2x-1} = 15\), both sides balanced perfectly, validating our solution. This step is not just a formality, but a necessary step in mathematical problem-solving as it ensures accuracy.

Isolating the Variable

To solve equations involving square roots, first aim to isolate the square root expression. This makes it easier to eliminate roots in later steps. In our problem:

• We began by subtracting 8 from both sides of the equation \(8 + \sqrt{2x-1} = 15\), leaving us with \(\sqrt{2x-1} = 7\).
• This simplification keeps the equation balanced, ensuring we maintain equality on both sides.

Such manipulation is vital because it sets up the problem for the next operation, such as squaring, which directly tackles the root.

Squaring Both Sides

Once the square root is isolated, the next step is to get rid of it through squaring. This involves squaring both sides of the equation to eliminate the root and simplify the expression:

• With \(\sqrt{2x-1} = 7\), squaring both sides transforms it into \((\sqrt{2x-1})^2 = 7^2\).
• This simplifies to \(2x - 1 = 49\), a linear equation that's solvable through basic operations.

Squaring both sides is a powerful method as it effectively removes the root, allowing for straightforward algebraic solutions in the subsequent steps. However, be mindful as sometimes squaring can introduce extraneous solutions, which is why checking is critical.