Solving Cubic Equations and Quadratics

The video tutorial explains how to solve an equation in quadratic form by recognizing the structure of the equation and using substitution. It demonstrates rewriting the equation in terms of a new variable, factoring the quadratic equation, and solving for the original variable. The process involves recognizing powers, performing substitutions, and using trial and error for factoring. The tutorial concludes with solving the factored equation to find the solutions for the original variable.

What indicates that the given equation is in quadratic form?

• The equation is cubic

• The equation is linear

• The presence of x raised to the power of 1/2

• The presence of x raised to the power of 1/3 and 2/3

What substitution is made to transform the equation into a quadratic form?

• Let u equal x squared

• Let u equal x

• Let u equal x raised to the power of 1/3

• Let u equal x raised to the power of 2/3

What is the first step in factoring the quadratic equation?

• Use the quadratic formula

• Simplify the equation

• Place the factors of 18u squared in the first positions

• Divide the equation by 2

Which method is used to factor the quadratic equation?

• Using the quadratic formula

• Trial and error

• Completing the square

• Graphing

What is done after factoring the quadratic equation in terms of u?

• Solve for u

• Substitute u back with x raised to the power of 1/3

• Graph the equation

• Use the quadratic formula

How do you solve for x after substituting back?

• Add 5 to both sides

• Cube both sides of the equation

• Divide by 3

• Subtract 5 from both sides

What is the result of cubing 5/3?

• 125/27

• 15/9

• 25/9

• 5/9

What is the solution for x when x raised to the power of 1/3 equals negative 5/6?

• x = -125/216

• x = 125/216

• x = -25/36

• x = 25/36

What is the purpose of raising both sides of the equation to the third power?

• To make the exponent equal to one

• To eliminate the fraction

• To simplify the equation

• To factor the equation

What are the final solutions to the original equation?

• x = 5/3 and x = -5/6

• x = 125/27 and x = -125/216

• x = 25/9 and x = -25/36

• x = 1 and x = -1