Solve the radical equation: [tex]$\sqrt{2 x-1}-3=0$[/tex]
If you have multiple answers separate them with a comma. If you have no solutions, write "no solutions"
Answer (2)
Isolate the radical term: 2 x − 1 = 3 .
Square both sides: 2 x − 1 = 9 .
Solve for x : x = 5 .
Check for extraneous solutions and confirm the solution: 5 .
Explanation
Understanding the Problem We are given the radical equation 2 x − 1 − 3 = 0 . Our goal is to solve for x . We need to isolate the radical term and then square both sides of the equation. Finally, we need to check for extraneous solutions.
Isolating the Radical First, we isolate the radical term by adding 3 to both sides of the equation: 2 x − 1 = 3
Squaring Both Sides Next, we square both sides of the equation to eliminate the square root: ( 2 x − 1 ) 2 = 3 2 2 x − 1 = 9
Solving for x Now, we solve for x by adding 1 to both sides: 2 x = 10 Then, we divide both sides by 2: x = 5
Checking for Extraneous Solutions We need to check if x = 5 is an extraneous solution by substituting it back into the original equation: 2 ( 5 ) − 1 − 3 = 10 − 1 − 3 = 9 − 3 = 3 − 3 = 0 Since the equation holds true, x = 5 is a valid solution.
Final Answer Therefore, the solution to the radical equation 2 x − 1 − 3 = 0 is x = 5 .
Examples
Radical equations are used in various fields, such as physics and engineering, to model real-world phenomena. For example, the period of a pendulum can be modeled using a radical equation. Solving such equations allows us to determine the length of the pendulum required to achieve a specific period. Similarly, in electrical engineering, radical equations can be used to analyze circuits and determine the values of components needed to achieve desired performance characteristics. Understanding how to solve radical equations is essential for solving practical problems in these fields.
The solution to the radical equation 2 x − 1 − 3 = 0 is x = 5 , which is confirmed by substituting back into the original equation. The steps involved isolating the radical, squaring both sides, solving for x , and checking for extraneous solutions. Therefore, there is one valid solution: x = 5 .
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