Quarter 1 - Illustrating and solving quadratic equations
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1) Quadratic equations are mathematical expressions of degree 2 that can be written in the standard form ax2 + bx + c = 0, where a, b, and c are real numbers and a ≠ 0. 2) There are four main methods for solving quadratic equations: extracting square roots, factoring, completing the square, and using the quadratic formula. 3) Factoring involves transforming the quadratic expression into standard form, factoring the expression into two linear factors, and setting each factor equal to 0 to solve for the roots.
Quarter 1 - Illustrating and solving quadratic equations
- 1. MATHEMATICS IX Quarter 1, week 1 Quadratic Equations REYNALDO M. ANARIO, JR. SubjectTeacher
- 2. Lesson 1 Objective: The Learner illustrates quadratic equations What is Quadratic Equation in one variable? Quadratic Equation in one variable is a mathematical sentence of degree 2 that can be written in the following standard form, 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0 Where a, b, and c are real numbers and a≠0
- 3. Examples: x2 – 4 = 0 y2 = 1 3a2 + 7a - 6 = 0 t + t2 = 6 25 = s2 32 = r2 3x(x – 2) = 10 When simplify 3x2 – 6x = 10 3x2 – 6x – 10 = 10 – 10 3x2 – 6x – 10 = 0 Standard form
- 4. Standard Form ax2 + bx + c = 0 ax 2 bx c Quadrati c Term Linear Term Constant Term x2 + 7x - 6 = 0 x 2 7x - 6 Quadrati c Term Linear Term Constant Term a = 1 b = 7 c = -6
- 5. The equation (2x + 5)(x – 1) = -6 is also a quadratic equation but is not written standard form. Transform into standard form (2x + 5)(x – 1) = -6 2x2 – 2x + 5x – 5 = -6 2x2 + 3x – 5 = -6 2x2 + 3x – 5 + 6 = -6 + 6 2x2 + 3x + 1 = 0 Standard Form 2x 2 3x 1 Quadrati c Term Linear Term Constant Term a = 2 b = 3 c = 1
- 6. The length of a rectangular lot is 15m longer than its width and the area is 100m2. Area of a rectangle A = Length x Width width = y length = y + 15 area = 100 100 = y(y + 15) 100 = y2 + 15y y2 + 15y = 100 y2 + 15y – 100 = 100 – 100 y2 + 15y – 100 = 0
- 7. Lesson 2 Objective: The Learner solves quadratic equation by: (a) extracting the square roots; (b) factoring; (c) completing the square; (d) using the quadratic formula How to solve quadratic equation? Solving Quadratic Equation is also finding the solution/s of the given quadratic equation. Solution/s of a quadratic equation is/are represented by the value of its variable.This value is called the root/s of a quadratic equation.
- 8. Solving Quadratic Equation b y Extracting Square Root Factoring Completing the Square Quadratic Formula
- 9. Solving Quadratic Equation b y Extracting Square Root Perfect Squares Roots 𝟒𝟗 − 𝟏𝟎𝟎 𝒙𝟐 𝒙𝟒 (𝒙 + 𝟐)𝟐 ±𝟕 −𝟏𝟎 𝒙 𝒙𝟐 𝒙 + 𝟐
- 10. Extracting Square Root REMEMBER 𝒙𝟐 = 𝒌, where k is the constant If k > 0, then 𝑥2 = 𝑘 has two real solutions or roots; 𝑥 = ± 𝑘 If k = 0, then 𝑥2 = 𝑘 has one real solution or root; 𝑥 = 0 If k < 0, then 𝑥2 = 𝑘 has no real solutions or
- 11. Extracting Square Root Example: Find the solutions of the equation 𝑥2 − 16 = 0 by extracting square roots. 𝒙𝟐 = 𝒌 𝒙𝟐 − 𝟏𝟔 = 𝟎 𝒙𝟐 − 𝟏𝟔 + 𝟏𝟔 = 𝟎 + 𝟏𝟔 𝒙𝟐 = 𝟏𝟔 𝒙𝟐 = ± 𝟏𝟔 𝒙 = ±𝟒 𝒙𝟏 = 𝟒 , 𝒙𝟐 = −𝟒
- 12. Extracting Square Root To check: Substitute these values (roots) in the original equation. 𝒙𝟐 − 𝟏𝟔 = 𝟎 𝒙𝟏 = 𝟒 𝒙𝟐 − 𝟏𝟔 = 𝟎 𝟒𝟐 − 𝟏𝟔 = 𝟎 𝟏𝟔 − 𝟏𝟔 = 𝟎 𝟎 = 𝟎 𝒙𝟐 = −𝟒 𝒙𝟐 − 𝟏𝟔 = 𝟎 (−𝟒)𝟐 −𝟏𝟔 = 𝟎 𝟏𝟔 − 𝟏𝟔 = 𝟎 𝟎 = 𝟎
- 13. Extracting Square Root Example: Find the solutions of the equation 𝑥 − 4 2 − 25 = 0. 𝒙𝟐 = 𝒌 𝒙 − 𝟒 𝟐 − 𝟐𝟓 = 𝟎 𝒙 − 𝟒 𝟐 − 𝟐𝟓 + 𝟐𝟓 = 𝟎 + 𝟐𝟓 𝒙 − 𝟒 𝟐 = 𝟐𝟓 𝒙 − 𝟒 𝟐 = ± 𝟐𝟓 𝒙 − 𝟒 = ±𝟓 𝒙𝟏 = 𝟓 + 𝟒 𝒙 − 𝟒 + 𝟒 = ±𝟓 + 𝟒 𝒙 = ±𝟓 + 𝟒 𝒙𝟐 = −𝟓 + 𝟒 𝒙𝟏 = 𝟗 𝒙𝟐 = −𝟏
- 14. Extracting Square Root To check: Substitute these values (roots) in the original equation. 𝒙𝟏 = 𝟗 (𝒙 − 𝟒)𝟐 −𝟐𝟓 = 𝟎 (𝟗 − 𝟒)𝟐 −𝟐𝟓 = 𝟎 (𝟓)𝟐 −𝟐𝟓 = 𝟎 𝟐𝟓 − 𝟐𝟓 = 𝟎 𝟎 = 𝟎 𝒙 − 𝟒 𝟐 − 𝟐𝟓 = 𝟎 𝒙𝟐 = −𝟏 (𝒙 − 𝟒)𝟐 −𝟐𝟓 = 𝟎 (−𝟏 − 𝟒)𝟐 −𝟐𝟓 = 𝟎 (−𝟓)𝟐 −𝟐𝟓 = 𝟎 𝟐𝟓 − 𝟐𝟓 = 𝟎 𝟎 = 𝟎
- 15. Solving Quadratic Equation b y Factoring Factorable expressions Factor/s 𝟐𝒙𝟐 − 𝟖𝒙 −𝟑𝒙𝟐 + 𝟗𝒙 𝒙𝟐 + 𝟖𝒙 + 𝟏𝟐 𝟐𝒙(𝒙 − 𝟒) −𝟑𝒙(𝒙 − 𝟑 (𝒙 + 𝟔)(𝒙 + 𝟐) 𝒙𝟐 + 𝟓𝒙 − 𝟔 (𝒙 + 𝟔)(𝒙 − 𝟏) 𝟐𝒙𝟐 − 𝟗𝒙 + 𝟕 (𝟐𝒙 − 𝟕)(𝒙 − 𝟏)
- 16. Factoring Step s: 1. Transform the quadratic equation into standard form if necessary. 2. Factor the quadratic expression. 3. Apply the zero product property by setting each factor of the quadratic expression equal to 0. 4. Solve each resulting equation. 5. Check the values of the variable obtained by substituting each in the original equation.
- 17. Factoring Examp le Find the solutions of the equation 𝑥2 + 8𝑥 = −15. 𝒙𝟐 + 𝟖𝒙 = −𝟏𝟓 𝒙𝟐 + 𝟖𝒙 + 𝟏𝟓 = −𝟏𝟓 + 𝟏𝟓 𝒙𝟐 + 𝟖𝒙 + 𝟏𝟓 = 𝟎 (𝒙 + 𝟓)(𝒙 + 𝟑) = 𝟎 𝒙 + 𝟓 = 𝟎 𝒙 + 𝟑 = 𝟎 𝒙 + 𝟓 − 𝟓 = 𝟎 − 𝟓 𝒙𝟏 = −𝟓 𝒙 + 𝟑 − 𝟑 = 𝟎 − 𝟑 𝒙𝟐 = −𝟑
- 18. Factoring To check: Substitute these values (roots) in the original equation. 𝒙𝟏 = −𝟓 (−𝟓)𝟐 +𝟖 −𝟓 = −𝟏𝟓 𝟐𝟓 − 𝟒𝟎 = −𝟏𝟓 −𝟏𝟓 = −𝟏𝟓 𝒙𝟐 + 𝟖𝒙 = −𝟏𝟓 𝒙𝟏 = −𝟑 (−𝟑)𝟐 +𝟖 −𝟑 = −𝟏𝟓 𝟗 − 𝟐𝟒 = −𝟏𝟓 −𝟏𝟓 = −𝟏𝟓
- 19. Factoring Examp le The length of a rectangular lot is 15m longer than its width and the area is 50m2. Find its dimensions. Area of a rectangle A = Length x Width width = y length = y + 15 area = 100 100 = y(y + 100 = y2 + 15y y2 + 15y = 100 y2 + 15y – 100 = 100 – 100 y2 + 15y – 100 =
- 20. Factoring Examp le The length of a rectangular lot is 15m longer than its width and the area is 100m2. Find its dimensions. 𝑦2 + 15𝑦 – 100 = 0 𝑦 + 20 𝑦 − 5 = 0 𝑦 + 20 = 0 𝑦 − 5 = 0 𝑦 + 20 − 20 = 0 − 20 𝑦1 = −20 𝑦 − 5 + 5 = 0 + 5 𝑦1 = 15 width= 𝑦 length= 𝑦 + 5 length= 15 + 5 length= 20𝑚 width= 15𝑚