MODULE 4

• DISCIPLINE • GOOD TASTE • EXCELLENCE
• Writer: Daren V. Perez
• Cover Illustrator: Joel J. Estudillo
• Department of Education National Capital Region SCHOOLS DIVISION OFFICE MARIKINA CITY
• MATHEMATICS Quarter 1: Module 4 Solving Equations Transformable to Quadratic Equation including Rational Algebraic Equations
• Hello Grade 9 learners! In this module, you will learn how to: Solve equations transformable to quadratic equation including rational algebraic equations.
• You can say that you have understood the lesson in this module if you can already:

  • Transform the algebraic equations into the standard form, 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0
  • Transform rational algebraic equations into quadratic equation in standard form, 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0
  • Find the solutions of the quadratic equations
  • Solve quadratic equations that are not written in standard form
  • Solve rational algebraic equations transformable to quadratic equation
• LESSON 1: SOLVING QUADRATIC EQUATIONS THAT ARE NOT WRITTEN IN STANDARD FORM
• Lovely wants to find out the dimensions of her room which is shown in the figure below.
• In solving quadratic equation that is not written in the standard form, transform the equation in the standard form 𝑎𝑥 2 + 𝑏𝑥 + �� = 0 where 𝑎, 𝑏, 𝑐 𝑎𝑟𝑒 𝑟𝑒𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟𝑠 𝑎𝑛𝑑 𝑎 ≠ 0 and then, solve the equation using any method in solving quadratic equation.
• Example 1: Solve 𝑥(𝑥 − 5) = 36.


  1. Simplify 𝑥(𝑥 − 5) = 36

  2. Apply Distributive Property 𝑥2 − 5𝑥 − 36 = 36 − 36

  3. Apply Addition Property of Equality 𝑥2 − 5𝑥 − 36 = 0

  4. Solve the equation using any method. Since the equation is factorable, then use factoring

  5. (�� − 9)(𝑥 + 4) = 0

  6. 𝑥 − 9 = 0 or 𝑥 + 4 = 0

  7. 𝑥 = 9 or 𝑥 = −4
• Example 2: Solve (𝑥 + 5)2 + (𝑥 − 2)2 = 37.


  1. Simplify (𝑥 + 5)2 + (𝑥 − 2)2 = 37

  2. Get the square of the two binomials (𝑥 + 5) and (𝑥 − 2) using FOIL method

  3. 𝑥2 + ��2 + 10𝑥 − 4�� + 25 + 4 − 37 = 0

  4. Combine like terms and apply Addition Property of Equality

  5. 2𝑥2 + 6𝑥 − 8 = 0

  6. Simplify the equation by dividing the equation by GCF 2.

  7. 𝑥2 + 3𝑥 − 4 = 0

  8. Transpose the constant term to right side of the equation.

  9. 𝑥2 + 3𝑥 = 4

  10. Get the half of 3 and then square it. Then add the result on both sides of the equation.

  11. 𝑥 + 3

  2 2 = 16 + 9

  4

  12. Factor the perfect square trinomial (𝑥 + 3

  1. 2 = 16 + 9

  4

  13. Extract the square root

  14. 𝑥 + 3

  2 = ± 5

  2

  15. Solve for 𝑥

  16. 𝑥1 = 2 or 𝑥2 = −4
• Example 3: Solve 2𝑥2 − 5𝑥 = 𝑥2 + 14.

  1. Simplify 2𝑥2 − 5𝑥 = 𝑥2 + 14

  2. Combine like terms and Apply Addition Property of Equality

  3. 𝑥2 − 5𝑥 − 14 = 0

  4. Identify the values of 𝑎, 𝑏 𝑎𝑛𝑑 𝑐

  5. Apply the quadratic formula

  6. 𝑥 = 5 ± 9

  2 = 7 or 𝑥 = −2
• Solving quadratic equations not in standard form

  1. Multiply both sides by the LCM of the denominators

  2. Write the resulting equation in standard form

  3. Solve the equation using any method (e.g. factoring)

  4. Check the solution by substituting back into the original equation
• There are rational equations that can be transformed into quadratic equations of the form ax^2 + bx + c = 0 where a, b and c are real numbers and a ≠ 0
• These transformed quadratic equations can be solved using methods for solving quadratic equations
• Equation in standard form

  x^2 + 10x - 144 = 0
• Solving the equation for the length of the deck

  1. Solve the quadratic equation using factoring or quadratic formula

  2. The solutions are the length and width of the deck
• The length and width of the rectangular deck can be found by solving the quadratic equation in standard form
• Solving rational algebraic equations

  1. Multiply both sides by LCM of denominators

  2. Combine like terms

  3. Write resulting quadratic equation in standard form

  4. Solve quadratic equation
• Rational algebraic equations can be transformed into quadratic equations
• Extraneous root/solution


  A solution of the equation derived from the original equation, but not a solution to the original equation
• Equations with extraneous roots/solutions do not have the extraneous root/solution as a valid solution to the original equation
• Rational algebraic equations transformable to quadratic equations

  • 𝑥 + 8/(𝑥-2) = 1 + 4𝑥/(𝑥-2)
  • 1/𝑥 + 2/(𝑥+1) = 3
  • 2/(𝑥-2) = 1/(𝑥+2) + 1
• Transforming rational algebraic equations to quadratic equations involves finding the LCM of the denominators, combining like terms, and writing the resulting equation in standard quadratic form
• Solving the quadratic equation can be done using methods like the quadratic formula or factoring
• Checking the solutions by substituting back into the original rational algebraic equation is an important step
• Extraneous roots/solutions can occur when transforming rational algebraic equations to quadratic equations
• Rational algebraic equation


  An equation that can be transformed into a quadratic equation
• Equation 1


  x + 8/5 + 2/x = x + 1
• Equation 2


  3/(x - 1) + 3x/(x - 5) = 1
• Equation 3

  (5x + 3x - 7)/x = (x + 1)/2