Solving Radical Equations and Extraneous Solutions

This video tutorial covers solving radical equations, starting with an introduction to radical equations and their components. It explains the steps to solve these equations, including isolating the radical, squaring both sides, solving the resulting equation, and checking for extraneous solutions. The tutorial provides several example problems, gradually increasing in complexity, and includes advanced examples involving quadratics. The video emphasizes the importance of checking for extraneous solutions and understanding the nature of radicals.

What is the first step in solving a radical equation?

• Multiply both sides by the radicand

• Square both sides of the equation

• Isolate the radical

• Add a constant to both sides

Why do we square both sides of a radical equation?

• To simplify the equation

• To isolate the variable

• To check for extraneous solutions

• To eliminate the radical

In the example where the equation is √(x-4) = 5, what is the value of x?

• 29

• 25

• 21

• 33

What should you do after solving the equation to ensure the solution is correct?

• Square both sides again

• Check for extraneous solutions

• Multiply both sides by the radicand

• Add a constant to both sides

In the example where the equation is 3√(x-7) - 9 = 6, what is the value of x?

• 35

• 32

• 29

• 25

What is an extraneous solution?

• A solution that does not satisfy the original equation

• A solution that is a negative number

• A solution that is a complex number

• A solution that satisfies the original equation

In the example where the equation is √(x-1) = √(5x-13), what is the value of x?

• 1

• 4

• 2

• 3

What should you do if you encounter a quadratic equation while solving a radical equation?

• Check for extraneous solutions

• Isolate the radical

• Set the equation to zero and factor

• Square both sides

In the quadratic example where the equation is x^2 - 7x + 6 = 0, what are the solutions?

• x = 2 and x = 3

• x = 3 and x = 6

• x = 1 and x = 6

• x = 1 and x = 7

In the final example, why were both solutions extraneous?

• Because they were positive

• Because they did not satisfy the original equation

• Because they were negative

• Because they were complex numbers