Finding and Estimating Square Roots ALGEBRA 1 LESSON 10-3 (For help, go to Lessons 1-2 and 8-5.) Simplify each expression. 1. 112 2. (–12)2 3. –(12)2 4. 1.52 5. 0.62 6. 2 7. 2 8. 2 1 2 3 – 4 5 10-3

Finding and Estimating Square Roots ALGEBRA 1 LESSON 10-3 Solutions 1. 112 = 11 • 11 = 121 2. (–12)2 = (–12)(–12) = 144 3. (–12)2 = –(12)(12) = –144 4. 1.52 = (1.5)(1.5) = 2.25 5. 0.62 = (0.6)(0.6) = 0.36 6. 2 = • = 7. 2 = = 8. 2 = • = 1 2 1 2 1 2 1 4 2 3 – 2 3 – 2 3 – 4 9 4 5 4 5 4 5 16 25 10-3

Finding and Estimating Square Roots ALGEBRA 1 LESSON 10-3 Simplify each expression. a. 25 positive square root = 5 b. ± 9 25 3 5 The square roots are and – . = ± c. – 64 negative square root = –8 d. –49 For real numbers, the square root of a negative number is undefined. is undefined e. ± 0 There is only one square root of 0. = 0 10-3

Finding and Estimating Square Roots ALGEBRA 1 LESSON 10-3 Tell whether each expression is rational or irrational. a. ± 144 = ± 12 rational b. – = –0.44721359 . . . 1 5 irrational c. – 6.25 = –2.5 rational d. = 0.3 1 9 rational e. 7 = 2.64575131 . . . irrational 10-3

Finding and Estimating Square Roots ALGEBRA 1 LESSON 10-3 Between what two consecutive integers is 28.34 ? 28.34 is between the two consecutive perfect squares 25 and 36. 25 < 28.34 < 36 The square roots of 25 and 36 are 5 and 6, respectively. 28.34 5 < < 6 28.34 is between 5 and 6. 10-3

Finding and Estimating Square Roots ALGEBRA 1 LESSON 10-3 Find 28.34 to the nearest hundredth. 28.34 5.323532662 Use a calculator. 5.32 Round to the nearest hundredth. 10-3

Finding and Estimating Square Roots ALGEBRA 1 LESSON 10-3 Suppose a rectangular field has a length three times its width x. The formula d = x2 + (3x)2 gives the distance of the diagonal of a rectangle. Find the distance of the diagonal across the field if x = 8 ft. d = x2 + (3x)2 d = 82 + (3 • 8)2 Substitute 8 for x. Simplify. d = 64 + 576 d = 640 Use a calculator. Round to the nearest tenth. d 25.3 The diagonal is about 25.3 ft long. 10-3

Solving Quadratic Equations ALGEBRA 1 LESSON 10-4 (For help, go to Lesson10-3.) Simplify each expression. 1. 36 2. – 81 3. ± 121 4. 1.44 5. 0.25 6. ± 1.21 7. 8. ± 9. ± 1 4 9 49 100 10-4

Solving Quadratic Equations ALGEBRA 1 LESSON 10-4 Solutions 1. 36 = 6 2. – 81 = –9 3. ± 121 = ±11 4. 1.44 = 1.2 5. 0.25 = 0.5 6. ± 1.21 = ±1.1 7. = 8. ± = ± 9. ± = 1 4 1 2 1 9 1 3 49 100 7 10 10-4

Solving Quadratic Equations ALGEBRA 1 LESSON 10-4 Solve each equation by graphing the related function. a. 2x2 = 0 b. 2x2 + 2 = 0 c. 2x2 – 2 = 0 Graph y = 2x2 Graph y = 2x2 + 2 Graph y = 2x2 – 2 There is one solution, x = 0. There is no solution. There are two solutions, x = ±1. 10-4

Solving Quadratic Equations ALGEBRA 1 LESSON 10-4 Solve 3x2 – 75 = 0. 3x2 – 75 + 75 = 0 + 75 Add 75 to each side. 3x2 = 75 x2 = 25 Divide each side by 3. x = ± 25 Find the square roots. x = ± 5 Simplify. 10-4

Solving Quadratic Equations ALGEBRA 1 LESSON 10-4 A museum is planning an exhibit that will contain a large globe. The surface area of the globe will be 315 ft2. Find the radius of the sphere producing this surface area. Use the equation S = 4 r2, where S is the surface area and r is the radius. S = 4 r 2 315 = 4 r 2 Substitute 315 for S. Put in calculator ready form. 315 (4) = r 2 = r 2 315 (4) Find the principle square root. 5.00668588 r Use a calculator. The radius of the sphere is about 5 ft. 10-4

Solving Quadratic Equations ALGEBRA 1 LESSON 10-4 1. Solve each equation by graphing the related function. If the equation has no solution, write no solution. a. 2x2 – 8 = 0 b. x2 + 2 = –2 2. Solve each equation by finding square roots. a. m2 – 25 = 0 b. 49q2 = 9 3. Find the speed of a 4-kg bowling ball with a kinetic energy of 160 joules. Use the equation E = ms2, where m is the object’s mass in kg, E is its kinetic energy, and s is the speed in meters per second. ±2 no solution ±5 3 7 ± 1 2 about 8.94 m/s 10-4

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 = 2 2. – 9 = 4 3. 7q + 16 = –3 Factor each expression. 4. 2c2 + 29c + 14 5. 3p2 + 32p + 20 6. 4x2 – 21x – 18 a 8 10-5

Factoring to Solve Quadratic Equations ALGEBRA 1 LESSON 10-5 Solutions 1. 6 + 4n = 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 = –19 + 16 = –3 a 8 a 8 104 8 5 7 5 7 19 7 10-5

Factoring to Solve Quadratic Equations ALGEBRA 1 LESSON 10-5 Solutions (continued) 4. 2c2 + 29c + 14 = (2c + 1)(c + 14) Check: (2c + 1)(c + 14) = 2c2 + 28c + c + 14 = 2c2 + 29c + 14 5. 3p2 + 32p + 20 = (3p + 2)(p + 10) Check: (3p + 2)(p + 10) = 3p2 + 30p + 2p + 20 = 3p2 + 32p + 20 6. 4x2 – 21x – 18 = (4x + 3)(x – 6) Check: (4x + 3)(x – 6) = 4x2 – 24x + 3x – 18 = 4x2 – 21x – 18 10-5

Factoring to Solve Quadratic Equations ALGEBRA 1 LESSON 10-5 Solve (2x + 3)(x – 4) = 0 by using the Zero Product Property. (2x + 3)(x – 4) = 0 2x + 3 = 0 or x – 4 = 0 Use the Zero-Product Property. 2x = –3 Solve for x. x = – 3 2 or x = 4 Check: Substitute – for x. 3 2 Substitute 4 for x. (2x + 3)(x – 4) = 0 [2(– ) + 3](– – 4) 0 [2(4) + 3](4 – 4) 0 (0)(– 5 ) = 0 1 (11)(0) = 0 10-5

Factoring to Solve Quadratic Equations ALGEBRA 1 LESSON 10-5 Solve x2 + x – 42 = 0 by factoring. x2 + x – 42 = 0 (x + 7)(x – 6) = 0 Factor using x2 + x – 42 x + 7 = 0 or x – 6 = 0 Use the Zero-Product Property. x = –7 or x = 6 Solve for x. 10-5

Factoring to Solve Quadratic Equations ALGEBRA 1 LESSON 10-5 Solve 3x2 – 2x = 21 by factoring. 3x2 – 2x = 21 Subtract 21 from each side. (3x + 7)(x – 3) = 0 Factor 3x2 – 2x – 21. 3x + 7 = 0 or x – 3 = 0 Use the Zero-Product Property 3x = –7 Solve for x. x = – or x = 3 7 3 10-5

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 + 1 + 1 = x + 2 The length of the material = x + 3 + 1 + 1 = x + 5 Relate: length  width = area of the sheet 10-5

Factoring to Solve Quadratic Equations ALGEBRA 1 LESSON 10-5 (continued) Write: (x + 2) (x + 5) = 130 x2 + 7x + 10 = 130 Find the product (x + 2) (x + 5). x2 + 7x – 120 = 0 Subtract 130 from each side. (x – 8) (x + 15) = 0 Factor x2 + 7x – 120. x – 8 = 0 or x + 15 = 0 Use the Zero-Product Property. x = 8 or x = –15 Solve for x. The only reasonable solution is 8. So the dimensions of the box are 8 in.  11 in.  1 in. 10-5

Factoring to Solve Quadratic Equations ALGEBRA 1 LESSON 10-5 1. Solve (2x – 3)(x + 2) = 0. Solve by factoring. 2. 6 = a2 – 5a 3. 12x + 4 = –9x2 4. 4y2 = 25 –2, 3 2 – 2 3 ± 5 2 –1, 6 10-5

Completing the Square Find each square. ALGEBRA 1 LESSON 10-6 (For help, go to Lessons 9-4 and 9-7.) Find each square. 1. (d – 4)2 2. (x + 11)2 3. (k – 8)2 Factor. 4. b2 + 10b + 25 5. t2 + 14t + 49 6. n2 – 18n + 81 10-6

Completing the Square Solutions ALGEBRA 1 LESSON 10-6 1. (d – 4)2 = d2 – 2d(4) + 42 = d2 – 8d + 16 2. (x + 11)2 = x2 + 2x(11) + 112 = x2 + 22x + 121 3. (k – 8)2 = k2 – 2k(8) + 82 = k2 – 16k + 64 4. b2 + 10b + 25 = b2 + 2b(5) + 52 = (b + 5)2 5. t2 + 14t + 49 = t2 + 2t(7) + 72 = (t + 7)2 6. n2 – 18n + 81 = n2 – 2n(9) + 92 = (n – 9)2 Solutions 10-6

Using the Quadratic Formula ALGEBRA 1 LESSON 10-7 (For help, go to Lesson 10-6.) Find the value of c to complete the square for each expression. 1. x2 + 6x + c 2. x2 + 7x + c 3. x2 – 9x + c Solve each equation by completing the square. 4. x2 – 10x + 24 = 0 5. x2 + 16x – 36 = 0 6. 3x2 + 12x – 15 = 0 7. 2x2 – 2x – 112 = 0 10-7

Using the Quadratic Formula ALGEBRA 1 LESSON 10-7 Solutions 1. x2 + 6x + c; c = 2 = 32 = 9 2. x2 + 7x + c; c = 2 = 3. x2 – 9x + c; c = 2 = 4. x2 – 10x + 24 = 0; 5. x2 + 16x – 36 = 0; c = 2 = (–5)2 = 25 c = 2 = 82 = 64 x2 – 10x = –24 x2 + 16x = 36 x2 – 10x + 25 = –24 + 25 x2 + 16x + 64 = 36 + 64 (x – 5)2 = 1 (x + 8)2 = 100 (x – 5) = ±1 (x + 8) = ±10 x – 5 = 1   or x – 5 = –1 x + 8 = 10  or x + 8 = –10 x = 6   or   x = 4 x = 2   or  x = –18 6 2 7 2 49 4 –9 2 81 4 –10 2 16 2 10-7

Using the Quadratic Formula ALGEBRA 1 LESSON 10-7 Solutions (continued) 6. 3x2 + 12x – 15 = 0 7. 2x2 – 2x – 112 = 0 3(x2 + 4x – 5) = 0 2(x2 – x – 56) = 0 x2 + 4x – 5 = 0; x2 – x – 56 = 0; c = 2 = 22 = 4 c = 2 = x2 + 4x = 5 x2 – x = 56 x2 + 4x + 4 = 5 + 4 x2 – x + = 56 + (x + 2)2 = 9 (x – )2 = (x + 2) = ±3 x – = ± x + 2 = 3   or x + 2 = –3 x – = or x – = x = 1   or   x = –5 x = 8 or x = –7 4 2 –1 2 1 4 1 4 1 4 1 2 225 4 1 2 15 2 1 2 15 2 1 2 –15 2 10-7

Using the Quadratic Formula ALGEBRA 1 LESSON 10-7 Solve x2 + 2 = –3x using the quadratic formula. x2 + 3x + 2 = 0 Add 3x to each side and write in standard form. x = –b ± b2 – 4ac 2a Use the quadratic formula. x = –3 ± (–3)2 – 4(1)(2) 2(1) Substitute 1 for a, 3 for b, and 2 for c. x = –3 ± 1 2 Simplify. x = –3 + 1 2 –3 – 1 or Write two solutions. x = –1 or x = –2 Simplify. 10-7

Using the Quadratic Formula ALGEBRA 1 LESSON 10-7 (continued) Check: for x = –1 for x = –2 (–1)2 + 3(–1) + 2 0 (–2)2 + 3(–2) + 2 0 1 – 3 + 2 0 4 – 6 + 2 0 0 = 0 10-7

Using the Quadratic Formula ALGEBRA 1 LESSON 10-7 Solve 3x2 + 4x – 8 = 0. Round the solutions to the nearest hundredth. x = –b ± b2 – 4ac 2a Use the quadratic formula. x = –4 ± 42 – 4(3)(–8) 2(3) Substitute 3 for a, 4 for b, and –8 for c. –4 ± 112 6 x = or Write two solutions. –4 + 112 6 –4 – 112 Use a calculator. x –4 + 10.583005244 6 –4 – 10.583005244 or x 1.10 x –2.43 Round to the nearest hundredth. 10-7

Using the Quadratic Formula ALGEBRA 1 LESSON 10-7 A child throws a ball upward with an initial upward velocity of 15 ft/s from a height of 2 ft. If no one catches the ball, after how many seconds will it land? Use the vertical motion formula h = –16t2 + vt + c, where h = 0, v = velocity, c = starting height, and t = time to land. Round to the nearest hundredth of a second. Step 1:  Use the vertical motion formula. h = –16t2 + vt + c 0 = –16t2 + 15t + 2  Substitute 0 for h, 15 for v, and 2 for c. Step 2: Use the quadratic formula. x = –b ± b2 – 4ac 2a 10-7

Using the Quadratic Formula ALGEBRA 1 LESSON 10-7 (continued) t = –15 ± 152 – 4(–16)(2) 2(–16) Substitute –16 for a, 15 for b, 2 for c, and t for x. –15 ± 225 + 128 –32 t = Simplify. –15 ± 353 t = –15 + 18.79 –32 or –15 – 18.79 Write two solutions. t –0.12 or 1.06 Simplify. Use the positive answer because it is the only reasonable answer in this situation. The ball will land in about 1.06 seconds. 10-7

Using the Quadratic Formula ALGEBRA 1 LESSON 10-7 Which method(s) would you choose to solve each equation? Justify your reasoning. a. 5x2 + 8x – 14 = 0 Quadratic formula; the equation cannot be factored easily. b. 25x2 – 169 = 0 Square roots; there is no x term. c. x2 – 2x – 3 = 0 Factoring; the equation is easily factorable. d. x2 – 5x + 3 = 0 Quadratic formula, completing the square, or graphing; the x2 term is 1, but the equation is not factorable. e. 16x2 – 96x + 135 = 0 Quadratic formula; the equation cannot be factored easily and the numbers are large. 10-7

Using the Quadratic Formula ALGEBRA 1 LESSON 10-7 1. Solve 2x2 – 11x + 12 = 0 by using the quadratic formula. 2. Solve 4x2 – 12x = 64. Round the solutions to the nearest hundredth. 3. Suppose a model rocket is launched from a platform 2 ft above the ground with an initial upward velocity of 100 ft/s. After how many seconds will the rocket hit the ground? Round the solution to the nearest hundredth. 1.5, 4 –2.77, 5.77 6.27 seconds 10-7