Factoring to Solve Quadratic Equations ALGEBRA 1 LESSON 10-5 (For help, go to Lessons 2-2 and 9-6.) Solve and check each equation. 1.6 + 4n = 22. – 9 =

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Factoring to Solve Quadratic Equations ALGEBRA 1 LESSON 10-5 (For help, go to Lessons 2-2 and 9-6.) Solve and check each equation n = 22. – 9 = 43.7q + 16 = –3 Factor each expression. 4.2c c p p x 2 – 21x – 18 a8a8 5-13

Factoring to Solve Quadratic Equations ALGEBRA 1 LESSON 10-5 Solutions n = 2 4n = –4 n = –1 Check: 6 + 4(–1) = 6 + (–4) = 2 2. – 9 = 4 = 13 a = 104 Check: – 9 = 13 – 9 = 4 3.7q + 16 = –3 7q = –19 q = –2 Check: 7 (–2 ) + 16 = 7(– ) + 16 = – = –3 a8a8 a8a

Factoring to Solve Quadratic Equations ALGEBRA 1 LESSON 10-5 Solutions (continued) 4.2c c + 14 = (2c + 1)(c + 14) Check: (2c + 1)(c + 14) = 2c c + c + 14 = 2c c p p + 20 = (3p + 2)(p + 10) Check: (3p + 2)(p + 10) = 3p p + 2p + 20 = 3p p x 2 – 21x – 18 = (4x + 3)(x – 6) Check: (4x + 3)(x – 6) = 4x 2 – 24x + 3x – 18 = 4x 2 – 21x –

Factoring to Solve Quadratic Equations ALGEBRA 1 LESSON 10-5 The diagram shows a pattern for an open-top box. The total area of the sheet of materials used to make the box is 130 in. 2. The height of the box is 1 in. Therefore, 1 in.  1 in. squares are cut from each corner. Find the dimensions of the box. Define: Let x = width of a side of the box. Then the width of the material = x = x + 2 The length of the material = x = x + 5 Relate: length  width = area of the sheet 5-13

Factoring to Solve Quadratic Equations ALGEBRA 1 LESSON 10-5 (continued) Write: (x + 2) (x + 5) = 130 x 2 + 7x + 10 = 130Find the product (x + 2) (x + 5). x 2 + 7x – 120 = 0Subtract 130 from each side. (x – 8) (x + 15) = 0Factor x 2 + 7x – 120. x – 8 = 0orx + 15 = 0Use the Zero-Product Property. x = 8or x = –15Solve for x. The only reasonable solution is 8. So the dimensions of the box are 8 in.  11 in.  1 in. 5-13

Factoring to Solve Quadratic Equations ALGEBRA 1 LESSON Solve (2x – 3)(x + 2) = 0. Solve by factoring. 2.6 = a 2 – 5a3.12x + 4 = –9x 2 4.4y 2 = 25 –2, 3232 –1, 6 – 2323 ±