MATH 9 - Solving Equations Transformable to Quadratic Equation Including Rational Algebraic Equation

Summary

In this educational video from 'Cup of Teach', viewers are guided through solving equations that can be transformed into quadratic equations, such as rational algebraic equations. The presenter demonstrates how to rewrite and simplify higher-degree equations into a quadratic form using substitution and factoring techniques. Examples include transforming x^4 - 5x^2 + 4 = 0 into a quadratic equation and solving it by factoring. The video also covers solving equations with rational expressions and provides solutions for each example, concluding with the roots of the equations.

Takeaways

• 📘 The video focuses on solving equations that can be transformed into quadratic equations, specifically rational algebraic equations.
• 🔄 The process involves rewriting the given equation x^4 - 5x^2 + 4 = 0 in a form that allows substitution to simplify it into a quadratic equation.
• 🆕 The substitution method is used by introducing a new variable 'a' to represent x^2, transforming the equation into a^2 - 5a + 4 = 0.
• ✅ Factoring is a key step in solving the quadratic equation, where the factors of 4 that sum to -5 are identified as -4 and -1.
• 🔍 The solutions for 'a' are found by setting each factor equal to zero, yielding a = 4 and a = 1.
• 🔄 The final step involves substituting back 'a' with x^2 to find the solutions for x, resulting in x = ± 2 and x = ± 1.
• 📐 The video also tackles a rational algebraic equation and uses the LCD (Least Common Denominator) method to clear the fractions.
• 🔢 After clearing the fractions, the equation is simplified and rearranged into a standard quadratic form, x^2 + 4x - 12 = 0.
• 🔄 The quadratic formula is applied to find the solutions for x, which are x = -6 and x = 2.
• 🔄 Another example involves transforming a nested radical equation into a quadratic equation by substituting x^2 - 4 with 't' and solving for 't'.
• 🔎 The solutions for 't' are 3 and -2, which are then substituted back to find the solutions for x, resulting in x = ± √7 and x = ± √2.

Q & A

What is the main topic of the 'Cup of Teach' video?

The main topic of the 'Cup of Teach' video is solving equations that are transformable to quadratic equations, including rational algebraic equations.

What is the first example equation given in the video?

The first example equation given in the video is x^4 - 5x^2 + 4 = 0.

How does the video suggest rewriting the first example equation?

The video suggests rewriting the first example equation by considering x^2 as a single entity, denoted by the variable 'a', transforming it into a^2 - 5a + 4 = 0.

What is the method used to factor the transformed equation in the video?

The method used to factor the transformed equation is finding two numbers that multiply to the constant term (4) and add up to the coefficient of the linear term (-5).

What are the solutions for 'a' after factoring the equation in the video?

The solutions for 'a' are a = 4 and a = 1.

How does the video relate the solutions for 'a' back to the original variable 'x'?

The video relates the solutions for 'a' back to the original variable 'x' by substituting 'a' with x^2, leading to the solutions x = ± 2 and x = ± 1.

What is the second type of equation transformation demonstrated in the video?

The second type of equation transformation demonstrated in the video is solving a rational algebraic equation by multiplying both sides by the least common denominator (LCD) and then simplifying.

What is the final form of the equation after multiplying by the LCD in the second example?

The final form of the equation after multiplying by the LCD in the second example is x^2 + 4x - 12 = 0.

What are the solutions to the second example equation?

The solutions to the second example equation are x = -6 and x = 2.

How does the video approach the third type of equation transformation?

The video approaches the third type of equation transformation by introducing a new variable 't' for the grouped terms, transforming it into a quadratic equation in terms of 't', and then solving for 't'.

What are the final solutions for 'x' in the third example after substituting back 't'?

The final solutions for 'x' in the third example are x = ± √7 and x = ± √2.