completing-the-square quadratic equation.pptx

completing-the-square quadratic equation.pptx

1. Solving Quadratic Equations by Completing the Square by: Teacher Novie Quarter 1 – Week 2 (Lesson 3) MELC Based GRADE 9 MATHEMATICS

2. LEARNING OBJECTIVE: 1. Solve quadratic equations by completing the square.

3. Perfect Square trinomial x 2  + 2 x  + 1 Square of Binomial (x  + 1)² x 2  + 20 x  + 100  (x  + 10)²   x 2   + 6 x  + 9 (x  + 3)²

4. Another method of solving quadratic equations is by completing the square . This method involves transforming the quadratic equation ax 2 + bx + c = 0 . into the form ( x – h) 2 = k , where k ≥ 0 . The value of k should be positive to obtain a real number solution.

5. 1. Divide both sides of the equation by a then simplify. 2. Write the equation such that the terms with variables are on the left side of the equation and the constant term is on the right side. 3. Add the square of one-half of the coefficient of x on both sides of the resulting equation. The left side of the equation becomes a perfect square trinomial. 4. Express the perfect square trinomial on the left side of the equation as a square of a binomial. 5. Solve the resulting quadratic equation by extracting the square root. 6. Solve the resulting linear equations. 7. Check the solutions obtained against the original equation. To solve the quadratic equation ax 2 + bx + c = 0 by completing the square, the following steps can be followed:

6. Example #1. Find the solutions of 2x 2 + 12x – 14 = 0 by completing the square . Solution: 2x 2 + 12x – 14 = 0 1. Divide both sides of the equation by the coefficient a then simplify. 2 . Rewrite the equation x 2 + 6x = 7 x 2 + 6x + 9 = 7 + 9 x 2 + 6x + 9 = 16 (x + 3) 2 = 16 4. Express the perfect square trinomial on the left side of the equation as a square of a binomial. 5. Solve the resulting quadratic equation by extracting the square x 2 + 6x - 7= 0

7. 6. Solve the resulting linear equations. x + 3 = 4 x + 3 = -4 x = 4 - 3 x = 1 x = -4 - 3 x = -7 Checking: 2x 2 + 12x – 14 = 0 For x = 1: For x = -7: 2x 2 + 12x – 14 = 0

8. Example #2. Find the solutions of x 2 - 8x - 9 = 0. Solution: x 2 - 8x - 9 = 0 x 2 - 8x = 9 x 2 - 8x + 16 = 9 + 16 x 2 - 8x + 16 = 25 (x - 4) 2 = 25

9. Checking: For x = 9: For x = -1:

10. Example #3. Find the solutions of x 2 – 2x = 7 by completing the square . Solution: x 2 – 2x = 7 ½(b)= 1/2(2) = 1² = 1 x 2 – 2x + 1 = 7 + 1 x 2 – 2x + 1 = 8 (x – 1) 2 = 8 x – 1 =   x – 1 =   x – 1 =  

11. Continuation…. x – 1 =   x – 1 = -   x =   x =