Chapter 10 Sec 3 Completing the Square

2 of 19 Algebra 1 Chapter 10 Sections 3 & 4 Use Square Root Property Solve x x + 25 = 49. First & Last term perfect Squares? Middle term = 2 x x x 5? Then (x + 5)(x + 5) = x x + 25 So, (x + 5) 2 = 49 Sq Rt Prop. x + 5 = 7 and x + 5 = –7 x = 2 and x = –12 {2, –12} Yes YesYes

3 of 19 Algebra 1 Chapter 10 Sections 3 & 4 Completing the Square To solve the previous equation, the quadratic expression on one side MUST be a perfect square. However few quadratic expressions are perfect squares. So we will use the following method to make a perfect square.

4 of 19 Algebra 1 Chapter 10 Sections 3 & 4 Completing the Square c =. 3.The trinomial now factors to x 2 + bx + c

5 of 19 Algebra 1 Chapter 10 Sections 3 & 4 Complete the Square Find the value of c that makes x x + c, then factor. x x + 36, c = 36 x x + 36, c = 36 (x + 6) 2 (x + 6) 2 1.Find b =, b/2 = … 2.c = (b/2) 2 3.Factor to (x + b/2) 2

6 of 19 Algebra 1 Chapter 10 Sections 3 & 4 Solving by Completing the Square 1.Put all x’s on one side and all numbers on the other. 2.If a = 1, divide each term by a 3. 4.Add to both sides of the equation. 5.The factor trinomial to 6.Use the Square Root Property 7.Solve for each root, both the ( + ) and the ( – ) x 2 + bx

7 of 19 Algebra 1 Chapter 10 Sections 3 & 4 Solve by Completing the Square Solve x 2 + 8x – 20 = x 2 + 8x = 20 x 2 + 8x = 20 x 2 + 8x + 16 = (x + 4) 2 = 36 x = – = 2 x = – 4 – 6 = – 10 x = – = 2 x = – 4 – 6 = – 10 {– 10, 2}

8 of 19 Algebra 1 Chapter 10 Sections 3 & 4 Solve by Completing the Square Solve x 2 – 14x + 3 = – 10 – 3 – 3 – 3 – 3 x 2 – 14x = – 13 x 2 – 14x = – 13 x 2 – 14x + 49 = – (x – 7) 2 = 36 {1, 13}

Chapter 10 Sec 4 Solving by Quadratic Formula

10 of 19 Algebra 1 Chapter 10 Sections 3 & 4 Quadratic Formula

11 of 19 Algebra 1 Chapter 10 Sections 3 & 4 Integral Roots x 2 – 2 x – 24 = 0 (x + 4)(x – 6) = 0 x + 4 = 0 x – 6 = 0 x = – 4 x = 6 Use two methods to solve x 2 – 2x – 24 = 0. First, factoring.

12 of 19 Algebra 1 Chapter 10 Sections 3 & 4 Two Rational Roots Second, Quadratic Equation x 2 – 2x – 24 = 0. a = 1, b = –2, c = – 24 {– 4, 6}

13 of 19 Algebra 1 Chapter 10 Sections 3 & 4 One Rational Roots a = 1, b = 22, c = 121 Solve x x +121 = 0. Identify a, b, and c. {– 11}

14 of 19 Algebra 1 Chapter 10 Sections 3 & 4 Two Irrational Roots a = 24, b = –14, c = – 6 Solve 24x 2 – 14x = 6. Rewrite in standard form 24x 2 – 14x – 6 = 0.

15 of 19 Algebra 1 Chapter 10 Sections 3 & 4Discriminant The discriminant can be used to determine the number and type of roots. The expression b 2 – 4ac is the discriminant.

16 of 19 Algebra 1 Chapter 10 Sections 3 & 4 Discriminant b 2 – 4ac.

17 of 19 Algebra 1 Chapter 10 Sections 3 & 4 Use the Discriminant a = 2, b = 10, c = 11 State the value of the discriminant. Then determine the number of real roots of the equation. a = 4, b = – 20, c = 25 a. 2x x + 11 = 0 b. 4x 2 – 20x + 25 = 0 b 2 – 4ac (– 20) 2 – 4(4)(25) 400 – 400 = 0 1 Real Root (10) 2 – 4(2)(11) 100 – 88 = 12 positive 2 Real Roots

18 of 19 Algebra 1 Chapter 10 Sections 3 & 4 Use the Discriminant a = 3, b = 4, c = 2 State the value of the discriminant. Then determine the number of real roots of the equation. c. 3x 2 + 4x = – 2 Rewrite: 3x 2 + 4x + 2 = 0 b 2 – 4ac (4) 2 – 4(3)(2) 16 – 24 = – 8 Negative No Real Roots

19 of 19 Algebra 1 Chapter 10 Sections 3 & 4 Daily Assignment Chapter 10 Sections 3 & 4 Study Guide (SG) Pg Odd