Problem 42 - College Algebra with Modeling and Visualization

Question

Solve the equation. Check your answers. $$ \sqrt{x}=\sqrt{x-5}+1 $$

Step-by-Step Solution

Answer

The solution is \( x = 9 \).

Step 1: Isolate the Square Root

To solve the equation \( \sqrt{x} = \sqrt{x-5} + 1 \), we first isolate the square root \( \sqrt{x-5} \) on one side. The equation already has \( \sqrt{x-5} \) isolated: \( \sqrt{x} - 1 = \sqrt{x-5} \).

Step 2: Square Both Sides

To eliminate the square roots, square both sides of the equation: \( \left( \sqrt{x} - 1 \right)^2 = (\sqrt{x-5})^2 \).Simplify both sides:\( x - 2\sqrt{x} + 1 = x - 5 \).

Step 3: Simplify the Equation

Subtract \( x \) from both sides:\( -2\sqrt{x} + 1 = -5 \).Subtract 1 from both sides:\( -2\sqrt{x} = -6 \).

Step 4: Solve for \( \sqrt{x} \)

Divide both sides by -2 to solve for \( \sqrt{x} \):\( \sqrt{x} = 3 \).

Step 5: Find \( x \)

Square both sides to find \( x \):\( x = 3^2 = 9 \).

Step 6: Check the Solution

Substitute \( x = 9 \) back into the original equation to verify:\( \sqrt{9} = \sqrt{9 - 5} + 1 \). Simplify both sides:\( 3 = 2 + 1 \).Since both sides are equal, \( x = 9 \) is indeed the solution.

Key Concepts

Solving Equations

Solving equations is a fundamental skill in college algebra. When faced with an equation involving variables, the objective is to determine the values of these variables that make the equation true. In our example, the equation is \( \sqrt{x} = \sqrt{x-5} + 1 \). To solve it, you must manipulate the equation to isolate the variable.

• Start by simplifying the equation, if possible. Here, it's already simplified but requires handling the square root terms properly.
• Next, we looked to isolate the more complex term \( \sqrt{x-5} \). This can involve moving terms across the equation's equal sign.
• In this case, subtraction was needed to isolate the square root: \( \sqrt{x} - 1 = \sqrt{x-5} \).

A good strategy is to eliminate radicals like square roots by squaring both sides, but always remember to retain balance in equations by performing the same operations on both sides.

Square Root

The square root function plays a significant role in algebra. In our equation \( \sqrt{x} = \sqrt{x-5} + 1 \), understanding how to handle square roots is key.

• The square root of a number \( a \) is a value that, when multiplied by itself, gives \( a \). It's denoted \( \sqrt{a} \).
• Managing equations with square roots often involves squaring the entire equation, as done in the solution.
• Squaring both sides of the equation removes the square root, making the equation easier to simplify and solve: \( \left( \sqrt{x} - 1 \right)^2 = (\sqrt{x-5})^2 \).

After squaring in our case, more algebraic manipulation helped isolate and eventually solve for the value of \( x \). Be cautious with squaring, as it can introduce extraneous solutions.

Algebraic Verification

Algebraic verification is an essential step to ensure the correctness of a solution. After finding a potential solution, as we did with \( x = 9 \), we substitute it back into the original equation to verify.

• Start by substituting the solution into the initial equation \( \sqrt{x} = \sqrt{x-5} + 1 \).
• Calculate each side of the equation with \( x = 9 \). The left side becomes \( \sqrt{9} = 3 \) and the right side becomes \( \sqrt{9 - 5} + 1 = 2 + 1 = 3 \).
• If both sides are equal, the solution is verified.

Verification ensures no errors occurred during solving and avoids incorrect solutions. This step concludes the solving process, providing confidence in the accuracy of the solution.