Problem 10 - Algebra for College Students

Question

Solve each equation. Don't forget to check each of your potential solutions. \(\sqrt{2 y-3}=5\)

Step-by-Step Solution

Answer

The solution is \( y = 14 \).

Step 1: Isolate the Radical

To solve the equation \(\sqrt{2y - 3} = 5\), we start by ensuring the radical (square root) is isolated. In this case, it already is, so no additional manipulation is needed.

Step 2: Eliminate the Radical

Square both sides of the equation to eliminate the square root. This gives: \( (\sqrt{2y - 3})^2 = 5^2 \).Simplifying this, we have: \[ 2y - 3 = 25 \].

Step 3: Solve for y

Now, solve for \(y\) by isolating it. Start by adding 3 to both sides of the equation:\[ 2y - 3 + 3 = 25 + 3 \]This simplifies to:\[ 2y = 28 \].Next, divide both sides by 2:\[ \frac{2y}{2} = \frac{28}{2} \]Resulting in:\[ y = 14 \].

Step 4: Check the Solution

To ensure \( y = 14 \) is correct, substitute it back into the original equation:\[ \sqrt{2(14) - 3} = \sqrt{28 - 3} = \sqrt{25} = 5 \].The left side equals the right side, so the solution is verified.

Key Concepts

Isolating the Radical

When solving an equation that includes a square root, the first step is to make sure the square root expression is by itself on one side of the equation. This is called isolating the radical. In the provided equation \( \sqrt{2y - 3} = 5 \), the radical \( \sqrt{2y - 3} \) is already isolated. This means the square root is on one side of the equation, with no other terms or coefficients. If it wasn't isolated, you would move any additional numbers or variables away by using basic algebraic operations (addition, subtraction, multiplication, or division). It's a crucial step because isolating the radical allows you to accurately perform the next steps, such as eliminating the square root.

Eliminating Square Roots

Eliminating a square root from an equation involves the process of squaring both sides. This action removes the square root symbol, simplifying the equation making it easier to solve. In our example, we square both sides of \( \sqrt{2y - 3} = 5 \). After squaring, the equation becomes:

• Square the left side: \((\sqrt{2y - 3})^2 = 2y - 3\)
• Square the right side: \(5^2 = 25\)

Thus, the resulting equation is \(2y - 3 = 25\), free of the square root. Squaring both sides is a reliable method, but it is important to remember that this step can introduce extraneous solutions, which must be verified later.

Checking Solutions

Checking solutions is a critical step in solving equations, especially those involving square roots. It ensures that the solutions obtained are valid for the original equation. Once we solve the equation and find \(y\), substituting back into the original equation verifies correctness. For \(y = 14\), substitute it back:

• Calculate the left side: \( \sqrt{2(14) - 3} = \sqrt{28 - 3} = \sqrt{25} = 5\)

Since the left side equals the right side, our solution is verified. If the sides did not match, it might indicate an extraneous solution—a result that emerged from manipulating the equation, such as through squaring—but isn't a true solution to the original problem.

Step-by-Step Solutions

Breaking down the solution process into clear, manageable steps can greatly aid understanding for complex problems. Here, the steps were:

• Isolate the Radical: Ensure the square root is on its own, which in this problem, it is already.
• Eliminate the Radical: Square both sides to get rid of the square root, leading to an equation that's easier to solve.
• Solve the Resulting Equation: Use algebraic techniques to solve for \(y\), giving \(y = 14\).
• Check the Solution: Verify by substituting \(y = 14\) back into the original equation to ensure it satisfies the mathematical condition.

This structured approach provides clarity and makes it easier to follow through and understand each step's purpose and execution.