Study Guide and Intervention. The Quadratic Formula and the Discriminant. Quadratic Formula. Replace a with 1, b with -5, and c with -14.

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Study Guide and Intervention Quadratic Formula The Quadratic Formula can be used to solve any quadratic equation once it is written in the form a 2 + b + c = 0. Quadratic Formula The solutions of a 2 + b + c = 0, with a 0, are given by = -b ± b2-4ac. 2a Eample Solve 2 - = 14 by using the Quadratic Formula. Rewrite the equation as 2 - - 14 = 0. -b ± b = 2-4ac 2a = -(-) ± (-) 2-4(1)(-14) 2(1) = ± 81 Simplify. 2 = ± 9 2 = 7 or -2 The solutions are -2 and 7. Quadratic Formula Replace a with 1, b with -, and c with -14. Solve each equation by using the Quadratic Formula. 1. 2 + 2-3 = 0 2. 2 + 10 + 24 = 0 3. 2-11 + 24 = 0 4. 4 2 + 19 - = 0. 14 2 + 9 + 1 = 0 6. 2 2 - - 1 = 0 7. 3 2 + = 2 8. 2y 2 + y - 1 = 0 9. 3 2-16 + 16 = 0 10. 8 2 + 6-9 = 0 11. r 2-3r + 2 2 = 0 12. 2-10 - 0 = 0 13. 2 + 6-23 = 0 14. 4 2-12 - 63 = 0 1. 2-6 + 21 = 0 Chapter 4 36 Glencoe Algebra 2

Study Guide and Intervention (continued) Roots and the Discriminant Discriminant Discriminant The epression under the radical sign, b 2-4ac, in the Quadratic Formula is called the discriminant. Type and Number of Roots Lesson b 2-4ac > 0 and a perfect square b 2-4ac > 0, but not a perfect square b 2-4ac = 0 b 2-4ac < 0 2 rational roots 2 irrational roots 1 rational root 2 comple roots Eample Find the value of the discriminant for each equation. Then describe the number and type of roots for the equation. a. 2 2 + + 3 The discriminant is b 2-4ac = 2-4(2) (3) or 1. The discriminant is a perfect square, so the equation has 2 rational roots. b. 3 2-2 + The discriminant is b 2-4ac = (-2) 2-4(3) () or -6. The discriminant is negative, so the equation has 2 comple roots. Complete parts a-c for each quadratic equation. a. Find the value of the discriminant. b. Describe the number and type of roots. c. Find the eact solutions by using the Quadratic Formula. 1. p 2 + 12p = -4 2. 9 2-6 + 1 = 0 3. 2 2-7 - 4 = 0 4. 2 + 4-4 = 0. 2-36 + 7 = 0 6. 4 2-4 + 11 = 0 7. 2-7 + 6 = 0 8. m 2-8m = -14 9. 2 2-40 = -16 10. 4 2 + 20 + 29 = 0 11. 6 2 + 26 + 8 = 0 12. 4 2-4 - 11 = 0 Chapter 4 37 Glencoe Algebra 2

Skills Practice Complete parts a-c for each quadratic equation. a. Find the value of the discriminant. b. Describe the number and type of roots. c. Find the eact solutions by using the Quadratic Formula. 1. 2-8 + 16 = 0 2. 2-11 - 26 = 0 3. 3 2-2 = 0 4. 20 2 + 7-3 = 0. 2-6 = 0 6. 2-6 = 0 7. 2 + 8 + 13 = 0 8. 2 - - 1 = 0 9. 2-2 - 17 = 0 10. 2 + 49 = 0 11. 2 - + 1 = 0 12. 2 2-3 = -2 Solve each equation by using the Quadratic Formula. 13. 2 = 64 14. 2-30 = 0 1. 2 - = 30 16. 16 2-24 - 27 = 0 17. 2-4 - 11 = 0 18. 2-8 - 17 = 0 19. 2 + 2 = 0 20. 3 2 + 36 = 0 21. 2 2 + 10 + 11 = 0 22. 2 2-7 + 4 = 0 23. 8 2 + 1 = 4 24. 2 2 + 2 + 3 = 0 2. PARACHUTING Ignoring wind resistance, the distance d(t) in feet that a parachutist falls in t seconds can be estimated using the formula d(t) = 16t 2. If a parachutist jumps from an airplane and falls for 1100 feet before opening her parachute, how many seconds pass before she opens the parachute? Chapter 4 38 Glencoe Algebra 2

Practice Solve each equation by using the Quadratic Formula. 1. 7 2 - = 0 2. 4 2-9 = 0 3. 3 2 + 8 = 3 4. 2-21 = 4 Lesson. 3 2-13 + 4 = 0 6. 1 2 + 22 = -8 7. 2-6 + 3 = 0 8. 2-14 + 3 = 0 9. 3 2 = -4 10. 2 2-20 - 6 = 0 11. 4 2-4 + 17 = 0 12. 8-1 = 4 2 13. 2 = 4-1 14. 4 2-12 + 7 = 0 Complete parts a-c for each quadratic equation. a. Find the value of the discriminant. b. Describe the number and type of roots. c. Find the eact solutions by using the Quadratic Formula. 1. 2-16 + 64 = 0 16. 2 = 3 17. 9 2-24 + 16 = 0 18. 2-3 = 40 19. 3 2 + 9-2 = 0 20. 2 2 + 7 = 0 21. 2-2 + 4 = 0 22. 12 2 - - 6 = 0 23. 7 2 + 6 + 2 = 0 24. 12 2 + 2-4 = 0 2. 6 2-2 - 1 = 0 26. 2 + 3 + 6 = 0 27. 4 2-3 2-6 = 0 28. 16 2-8 + 1 = 0 29. 2 2 - - 6 = 0 30. GRAVITATION The height h(t) in feet of an object t seconds after it is propelled straight up from the ground with an initial velocity of 60 feet per second is modeled by the equation h(t) = -16t 2 + 60t. At what times will the object be at a height of 6 feet? 31. STOPPING DISTANCE The formula d = 0.0s 2 + 1.1s estimates the minimum stopping distance d in feet for a car traveling s miles per hour. If a car stops in 200 feet, what is the fastest it could have been traveling when the driver applied the brakes? Chapter 4 39 Glencoe Algebra 2

Word Problem Practice 1. PARABOLAS The graph of a quadratic equation of the form y = a 2 + b + c is shown below. y 4. EXAMPLES Give an eample of a quadratic function f() that has the following properties. I. The discriminant of f is zero. II. There is no real solution of the equation f() = 10. Sketch the graph of = f(). - Is the discriminant b 2-4ac positive, negative, or zero? Eplain. O O y 2. TANGENT Kathleen is trying to find b so that the -ais is tangent to the parabola y = 2 + b + 4. She finds one value that works, b = 4. Is this the only value that works? Eplain. 3. SPORTS In 1990, American Randy Barnes set the world record for the shot put. His throw can be described by the equation y = 16 2 + 368. Use the Quadratic Formula to find how far his throw was to the nearest foot.. TANGENTS The graph of y = 2 is a parabola that passes through the point at (1, 1). The line y = m - m + 1, where m is a constant, also passes through the point at (1, 1). a. To find the points of intersection between the line y = m - m + 1 and the parabola y = 2, set 2 = m - m + 1 and then solve for. Rearranging terms, this equation becomes 2 - m + m - 1 = 0. What is the discriminant of this equation? b. For what value of m is there only one point of intersection? Eplain the meaning of this in terms of the corresponding line and the parabola. Chapter 4 40 Glencoe Algebra 2

Enrichment Sum and Product of Roots Sometimes you may know the roots of a quadratic equation without knowing the equation itself. Using your knowledge of factoring to solve an equation, you can work backward to find the quadratic equation. The rule for finding the sum and product of roots is as follows: Sum and Product of Roots If the roots of a 2 + b + c = 0, with a 0, are s 1 and s 2, then s 1 + s 2 = - a b and s s = c 1 2 a. Lesson Eample Write a quadratic equation that has the roots 3 and -8. The roots are = 3 and = -8. 3 + (-8) = - Add the roots. y 10 3(-8) = -24 Multiply the roots. Equation: 2 + - 24 = 0 8 6 4 2 O 10 2 4 20 ( 2, 30 1 4) 30 Write a quadratic equation that has the given roots. 1. 6, -9 2., -1 3. 6, 6 4. 4 ± 3. - 2, 2 7 Find k such that the number given is a root of the equation. 7. 7; 2 2 + k - 21 = 0 8. -2; 2-13 + k = 0 6. -2 ± 3 7 Chapter 4 41 Glencoe Algebra 2

Spreadsheet Activity Approimating the Real Zeros of Polynomials You have learned the Location Principle, which can be used to approimate the real zeros of a polynomial. 1 2 3 4 A B C f() D 23 E 3.3333333 9.1111111 F 1.6666667 0.7777778 G 0 2 H 1.6666667 0.7777778 I 3.3333333 9.1111111 J 23 Sheet 1 Sheet 2 Sheet 3 In the spreadsheet above, the positive real zero of ƒ() = 2-2 can be approimated in the following way. Set the spreadsheet preference to manual calculation. The values in A2 and B2 are the endpoints of a range of values. The values in D2 through J2 are values equally in the interval from A2 to B2. The formulas for these values are A2, A2 + (B2 - A2)/6, A2 + 2*(B2 - A2)/6, A2 + 3*(B2 - A2)/6, A2 + 4*(B2 - A2)/6, A2 + *(B2 - A2)/6, and B2, respectively. Row 3 gives the function values at these points. The function ƒ() = 2-2 is entered into the spreadsheet in Cell D3 as D2^2-2. This function is then copied to the remaining cells in the row. You can use this spreadsheet to study the function values at the points in cells D2 through J2. The value in cell F3 is positive and the value in cell G3 is negative, so there must be a zero between -1.6667 and 0. Enter these values in cells A2 and B2, respectively, and recalculate the spreadsheet. (You will have to recalculate a number of times.) The result is a new table from which you can see that there is a zero between 1.41414 and 1.414306. Because these values agree to three decimal places, the zero is about 1.414. This can be verified by using algebra. By solving 2-2 = 0, we obtain = ± 2. The positive root is = ± 2 = 1.414213..., which verifies the result. 1. Use a spreadsheet like the one above to approimate the zero of ƒ() = 3-2 to three decimal places. Then verify your answer by using algebra to find the eact value of the root. 2. Use a spreadsheet like the one above to approimate the real zeros of f() = 2 + 2 + 0.. Round your answer to four decimal places. Then, verify your answer by using the quadratic formula. 3. Use a spreadsheet like the one above to approimate the real zero of ƒ() = 3-3 2 2-6 - 2 between -0.4 and -0.3. Chapter 4 42 Glencoe Algebra 2