Problem 71 - Beginning and Intermediate Algebra

Question

Solve. $$\sqrt{2 y-5}-\sqrt{y-3}=1$$

Step-by-Step Solution

Answer

The short answer based on the given step-by-step solution is: The solution to the equation \(\sqrt{2y-5}-\sqrt{y-3}=1\) is \(y=7\) and \(y=3\) . Both of these solutions are valid when verified by substituting them back into the original equation.

Step 1: Isolate one square root term

We will isolate the square root term \(\sqrt{2y-5}\) on the left side by adding the other square root term \(\sqrt{y-3}\) to both sides of the equation: \[\sqrt{2y-5} = 1+\sqrt{y-3}\]

Step 2: Square both sides of the equation

To eliminate the square roots, we will square both sides of the equation: \[(\sqrt{2y-5})^2 = (1+\sqrt{y-3})^2\]

Step 3: Simplify the expression

Expand the squared terms and simplify: \[2y-5 = 1+2\sqrt{y-3}+y-3\]

Step 4: Solve for y

First, combine like terms by subtracting y from both sides: \[y-5 = 1+2\sqrt{y-3}-3\] Next, combine the constants on the right side and simplify: \[y-5 = -2+2\sqrt{y-3}\] Now, we will isolate the square root term on the right side by adding 2 to both sides: \[y-3 = 2\sqrt{y-3}\] To eliminate the square root, we will square both sides again: \[(y-3)^2 = (2\sqrt{y-3})^2\] Simplify the expressions: \[y^2-6y+9 = 4(y-3)\] Expand the expression on the right side and simplify: \[y^2-6y+9 = 4y-12\] Now, rearrange the terms to form a quadratic equation: \[y^2-10y+21=0\]

Step 5: Solve the quadratic equation

Using factoring, we can find the values of y that satisfy the equation: \[(y-7)(y-3)=0\] This gives us two possible values for y: \[y=7 \quad \text{or} \quad y=3\]

Step 6: Verify the solutions

Lastly, we need to verify our solutions by substituting them back into the original equation: \[\sqrt{2(7)-5}-\sqrt{7-3}=1\] \[\sqrt{9}-\sqrt{4}=1\] \[3-2=1\] \[1=1\] The first solution, y = 7, is valid. Now check the second solution: \[\sqrt{2(3)-5}-\sqrt{3-3}=1\] \[\sqrt{1}-\sqrt{0}=1\] \[1-0=1\] \[1=1\] Both solutions, y = 7 and y = 3, are valid.

Key Concepts

Solving Radical Equations

When solving radical equations, the main goal is to eliminate the radical (i.e., the square root) in the equation. This is usually done by squaring both sides to get rid of the square roots and simplify the equation.
Here's a basic strategy to solve a radical equation:

• Isolate one of the square root terms: Start by moving one radical to one side of the equation to make it easier to handle.
• Square both sides: Eliminate the square root by squaring each side of the equation. Remember, squaring is a powerful tool, but it must be applied carefully to ensure that each step is valid and doesn’t introduce extraneous solutions.
• Simplify and repeat if necessary: After squaring, simplify the resulting equation and, if needed, isolate any remaining radical. Repeat the process until no radicals are left.

Solve the new, simpler equation. The solution might lead to quadratic or linear equations that can be solved by known methods. Remember, after resolving the equation, it’s crucial to double-check the answers by plugging them back into the original equation to ensure they work. This process avoids accepting any extraneous solutions that have been introduced by squaring.

Quadratic Equations

Quadratic equations are often the result of solving radical equations, especially after squaring them multiple times to remove the radicals. A quadratic equation typically has the form: \[ ax^2 + bx + c = 0 \] where \( a \), \( b \), and \( c \) are constants, and \( x \) represents the unknown variable. There are several techniques to solve quadratic equations:

• Factoring: This involves expressing the quadratic equation as a product of two binomials. If the equation can be factored easily, it's a quick and efficient method.
• The Quadratic Formula: Use the formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) when factoring is not straightforward. This formula gives you the roots directly, but be sure to handle calculations carefully to find correct solutions.
• Completing the Square: Transforming the equation into a perfect square form can also be useful, especially for understanding the properties of the equation, such as finding the vertex of a parabola.

Always check the derived solutions against the original equation as it may contain terms or restrictions not present in the quadratic form. This is important in confirming the validity of the solution.

Verification of Solutions

Verification is an essential step in solving equations, especially when dealing with radicals or quadratics where the process might introduce extraneous solutions.
Follow this systematic approach:

• Substitute Solutions Back: Insert the solutions obtained into the original equation to verify if they truly satisfy it. With radical terms, calculate carefully to confirm.
• Check Each Step: During verification, ensure each solution logically follows from each manipulation made in the initial steps. If any steps involved squaring both sides, pay attention as this step could introduce solutions that do not actually satisfy the initial equation.

Keep in mind,

• Sometimes squaring an equation can lead to additional solutions that weren't present in the original equation.
• This makes verification a crucial step to validate that the solutions are not only mathematically correct but also relevant to the given problem.

In this way, you ensure the solutions are both numerically and contextually appropriate, reflecting the original problem's intentions without error.