DO NOT EDIT--Changes must be made through “File info” CorrectionKey=NL-A;CA-A Name Class Date 22.3 Using the q uadratic f ormula to Solve Equations Essential Question: What is the quadratic formula, and how can you use it to solve quadratic equations? Resource Locker Deriving the q uadratic f ormula Explore You can complete the square on the general form of a quadratic equation to derive a formula that can be used to solve any quadratic equation. A Write the standard form of a quadratic equation. ax 2 + bx + c = B 0 Subtract c from both sides. ax 2 + bx = -c C Multiply both sides by 4a to make the coefficient of x 2 a perfect square. 4a 2x 2 + D 4abx = -4ac Add b 2 to both sides of the equation to complete the square. © Houghton Mifflin Harcourt Publishing Company 4a 2x 2 + 4abx + b 2 = -4ac + b 2 E F Factor the left side to write the trinomial as the square of a binomial. ( 2ax + b 2 Take the square roots of both sides. __ 2ax + b G ) = b - 4ac 2 √ =± Subtract b from both sides. __ 2ax = -b ± H b 2 - 4ac √ b 2 - 4ac Divide both sides by 2a to solve for x. __ √ b 2 - 4ac -b ± __ x= Module 22 A1_MNLESE368187_U8M22L3.indd 1059 2a 1059 Lesson 3 3/23/14 6:30 AM DO NOT EDIT--Changes must be made through “File info” CorrectionKey=NL-A;CA-A DO NOT Correcti __ -b ± √b 2 - 4ac The formula you just derived, x = __, is called the quadratic formula. 2a It gives you the values of x that solve any quadratic equation where a ≠ 0. Reflect 1. What If? If the derivation had begun by dividing each term by a, what would the resulting binomial of x have been after completing the square? Does one derivation method appear to be simpler than the other? Explain. + bx + c = 0 b c 2 _ x +_ ax + a = 0 b c 2 _ x +_ ax = - a 2 2 b b b c+ _ 2 _ _ x +_ x + = a 2 2a a 2 2a b b c _ _ x+ = -a + 2a 2a b The resulting binomial is x + . The previous derivation appears simpler because no 2a fractions are involved in the derivation. 2 ax ( ( ) _) ( ) ( ) (_) Using the Discriminant to Determine the Number of Real Solutions Explain 1 Recall that a quadratic equation, ax 2 + bx + c, can have two, one, or no real solutions. By evaluating the part of the quadratic formula under the radical sign, b 2 - 4ac, called the discriminant, you can determine the number of real solutions. Example 1 Determine how many real solutions each quadratic equation has. x 2 - 4x + 3 = 0 Identify a, b, and c. b 2 - 4ac Use the discriminant. (-4) 2 - 4(1)(3) Substitute the identified values into the discriminant. 16 - 12 = 4 Simplify. © Houghton Mifflin Harcourt Publishing Company a = 1, b = -4, c = 3 Since b 2 - 4ac > 0, the equation has two real solutions. x 2 - 2x + 2 = 0 a = 1 , b = -2 , c = 2 Identify a, b, and c. b 2 - 4ac Use the discriminant. ( -2 ) - 4( 2 4 - 8 1 )( = -4 2 ) Substitute the identified values into the discriminant. Simplify. Since b 2 - 4ac < 0, the equation has no real solution(s). Module22 A1_MNLESE368187_U8M22L3.indd 1060 1060 Lesson3 3/23/14 6:30 AM DO NOT EDIT--Changes must be made through “File info” CorrectionKey=NL-A;CA-A Reflect 2. When the discriminant is positive, the quadratic equation has two real solutions. When the discriminant is negative, there are no real solutions. How many real solutions does a quadratic equation have if its discriminant equals 0? Explain. One real solution; if the discriminant is 0, then you are adding or subtracting the square root of 0 in the quadratic formula. Since the only square root of 0 is 0, the answer will be the same whether it is added or subtracted. Your Turn Use the discriminant to determine the number of real solutions for each quadratic equation. 3. x 2 + 4x + 1 = 0 4. 2x 2 - 6x + 15 = 0 a = 1, b = 4, c = 1 a = 2, b = -6, c = 15 (4) - 4(1)(1) (-6) - 4(2)(15) b 2 - 4ac b 2 - 4ac 2 2 16 - 4 = 12 two real solutions 36 - 120 = -84 no real solutions 5. x 2 + 6x + 9 = 0 a = 1, b = 6, c = 9 b 2 - 4ac (6) 2 - 4(1)(9) 36 - 36 = 0 one real solution Solving Equations by Using the Quadratic Formula Explain 2 To use the quadratic formula to solve a quadratic equation, check that the equation is in standard form. If not, rewrite it in standard form. Then substitute the values of a, b, and c into the formula. Example 2 Solve using the quadratic formula. © Houghton Mifflin Harcourt Publishing Company 2x 2 + 3x - 5 = 0 a = 2, b = 3, c = -5 Identify a, b, and c. -b ± √b - 4ac x = __ 2a Use the quadratic formula. __ 2 ___ 2 -3 ± √(3) - 4(2)(-5) x = ___ 2(2) _ Substitute the identified values into the quadratic formula. -3 ± √ 49 x=_ 4 -3 ± 7 x=_ 4 -3 + 7 -3 - 7 _ x= or x=_ 4 4 5 _ x = 1 or x=2 5. The solutions are 1 and -_ 2 Write as two equations. Module22 1061 A1_MNLESE368187_U8M22L3.indd 1061 Simplify the radicand and the denominator. Evaluate the square root. Simplify both equations. Lesson3 3/23/14 6:30 AM DO NOT EDIT--Changes must be made through “File info” CorrectionKey=NL-A;CA-A DO NOT Correcti Graph y =2 x 2 + 3x - 5 to verify your answers. The graph does verify the solutions. 2x = x 2 - 4 x 2 - 2x - 4 = 0 Write in standard form. a = 1 , b = -2 , c = -4 Identify a, b, and c. -b ± √b 2 - 4ac x = __ 2a Use the quadratic formula. ――― ――――――――― ( ) √( -2 ) - 4( 1 )( -4 ) x = ―――――――――――――― 2( 1 ) ―― - -2 ± 2 Substitute the identified values into the quadratic formula. √ 2± 20 x = __ 2 ――― √ ― Simplify the radicand and the denominator. ― 2± 2 ± 2 √5 4 ∙5 x = __ = _= 1 ± √5 2 2 Simplify. x = 1 + √5 or x = 1 - √5 Write as two equations. ― or x ≈ -1.236 The exact solutions are 3.236 ― ― Use a calculator to find approximate solutions to three decimal places. 1 + √5 and 1 - √5 . The approximate solutions are and -1.236 . Graph y = x 2 - 2x - 4 and find the zeros using the graphing calculator. The calculator will give approximate values. The graph does confirm the solutions. © Houghton Mifflin Harcourt Publishing Company x ≈ 3.236 ― Reflect 6. Discussion How can you use substitution to check your solutions? Substitute each value into the given quadratic equation to see if it leads to a true equality. Module22 A1_MNLESE368187_U8M22L3.indd 1062 1062 Lesson3 3/23/14 6:30 AM DO NOT EDIT--Changes must be made through “File info” CorrectionKey=NL-A;CA-A Your Turn Solve using the quadratic formula. 7. x 2 - 6x - 7 = 0 ―――――― -(-6)± √(-6) - 4(1)(-7) ___ x= 2(1) 6 ± √― 64 x=_ 2 6±8 _ x= 2 6+8 6-8 _ x= or x = _ 8. 2 2 x = 7 or 2x 2 = 8x - 7 ―――――― -(-8)± √(-8) - 4(2)(7) ___ x= 2(2) 8 ± √― 8 x=_ 4 8 ±2 √― 2 _ x= 4 ― 8 -2 √― 8 +2 √2 2 or x=_ x=_ 4 4 √― √― 2 2 or x = 2 - _ x=2+_ 2 2 √― √― 2 2 The solutions are 2 + _and 2 - _ . 2 2 x = -1 The solutions are 7 and -1. 2 Explain 3 2 Using the Discriminant with Real-World Models Given a real-world situation that can be modeled by a quadratic equation, you can find the number of real solutions to the problem using the discriminant, and then apply the quadratic formula to obtain the solutions. After finding the solutions, check to see if they make sense in the context of the problem. In projectile motion problems where the projectile height h is modeled by the equation h = −16t 2 + vt + s, where t is the time in seconds the object has been in the air, v is the initial vertical velocity in feet per second, and s is the initial height in feet. The -16 coefficient in front of the t 2 term refers to the effect of gravity on the object. This equation can be written using metric units as h = −4.9t 2 + vt + s, where the units are converted from feet to meters. Time remains in units of seconds. Example 3 For each problem, use the discriminant to determine the number of real © Houghton Mifflin Harcourt Publishing Company solutions for the equation. Then, find the solutions and check to see if they make sense in the context of the problem. A diver jumps from a platform 10 meters above the surface of the water. The diver’s height is given by the equation h = −4.9t 2 + 3.5t + 10, where t is the time in seconds after the diver jumps. For what time t is the diver’s height 1 meter? Substitute h = 1 into the height equation. Then, write the resulting quadratic equation in standard form to solve for t. 1 = −4.9t 2 + 3.5t + 10 0 = −4.9t 2 + 3.5t + 9 First, use the discriminant to find the number of real solutions of the equation. b 2 - 4ac Use the discriminant. (3.5)2 - 4(-4.9)(9) = 188.65 Since Module22 A1_MNLESE368187_U8M22L3.indd 1063 b 2 − 4ac > 0, the equation has two real solutions. 1063 Lesson3 3/23/14 6:30 AM DO NOT EDIT--Changes must be made through “File info” CorrectionKey=NL-A;CA-A DO NOT Correcti Next, use the quadratic formula to find the real number solutions. a = −4.9, b = 3.5, c = 10 Identify a, b, and c. ――― -b ± √b - 4ac 2 t = __ 2a -3.5 ± √188.65 t = __ 2(-4.9) Use the quadratic formula. ――― Substitute the identified values into the quadratic formula and the value of the discriminant. -3.5 ± 13.73 t ≈ __ -9.8 -3.5 - 13.73 -3.5 + 13.73 or t ≈ __ t ≈ __ -9.8 -9.8 t ≈ -1.04 or t ≈ 1.76 Simplify. Write as two equations. Solutions Disregard the negative solution because t represents the seconds after the diver jumps and a negative value has no meaning in this context. So, the diver is at height 1 meter after a time of t ≈ 1.76 seconds. B The height in meters of a model rocket on a particular launch can be modeled by the equation h = −4.9t 2 + 102t + 100, where t is the time in seconds after its engine burns out 100 meters above the ground. When will the rocket reach a height of 600 meters? Substitute h = 600 into the height equation. Then, write the resulting quadratic equation in standard form to solve for t. h = −4.9t 2 + 102t + 100 600 = −4.9t 2 + 102t + 100 0 = −4.9t 2 + 102t - 500 First, use the discriminant to find the number of real solutions of the equation. a = −4.9, b = 102 , c = -500 Identify a, b, and c. b 2 - 4ac Use the discriminant. 10404 - 9800 -500 ) = 604 Substitute the identified values into the discriminant. Simplify. Since b 2 - 4ac > 0, the equation has 2 real solutions. Module 22 A1_MNLESE368187_U8M22L3.indd 1064 1064 © Houghton Mifflin Harcourt Publishing Company ( 102 ) - 4(-4.9)( 2 Lesson 3 3/23/14 6:30 AM DO NOT EDIT--Changes must be made through “File info” CorrectionKey=NL-A;CA-A Next, use the quadratic formula to find the real number solutions. ――― √ 102 604 - ± t = __ 2(-4.9) Substitute the identified values into the quadratic formula and the value of the discriminant. -102 24.58 ± t = ___ -9.8 Simplify. 24.58 24.58 - 102 + - 102 t ≈ __ or t ≈ __ -9.8 -9.8 Write as two equations. t ≈ -7.90 Solutions Disregard the or negative launched and a negative t ≈ 12.92 solution because t represents the seconds after the rocket has value has no meaning in this context. So, the rocket is at height 600 meters after a time of t ≈ 12.92 seconds. Your Turn For each problem, use the discriminant to determine the number of real solutions for the equation. Then, find the solutions and check to see if they make sense in the context of the problem. 9. A soccer player uses her head to hit a ball up in the air from a height of 2 meters with an initial vertical velocity of 5 meters per second. The height h in meters of the ball is given by h = −4.9t 2 + 5t + 2, where t is the time elapsed in seconds. How long will it take the ball to hit the ground if no other players touch it? h = −4.9t 2 + 5t + 2 0 = −4.9t 2 + 5t + 2 © Houghton Mifflin Harcourt Publishing Company Find the discriminant. (5)2 − 4(-4.9)(2) = 64.2 Since b 2 - 4ac > 0, the equation has two real solutions. Use the quadratic formula to find the solutions of the quadratic equation. 64.2 -5 ± √―― __ -9.8 -5 ± 8.01 t≈_ t= -9.8 t ≈ -0.31 or t ≈ 1.33 Disregard the negative solution because there is no negative time in this problem context. The soccer ball reached the ground after about t ≈ 1.33 seconds. Module 22 A1_MNLESE368187_U8M22L3.indd 1065 1065 Lesson 3 3/23/14 6:30 AM DO NOT EDIT--Changes must be made through “File info” CorrectionKey=NL-A;CA-A DO NOT Correcti 10. The quarterback of a football team throws a pass to the team’s receiver. The height h in meters of the football can be modeled by h = −4.9t 2+ 3t + 1.75, where t is the time elapsed in seconds. The receiver catches the football at a height of 2 meters. How long does the ball remain in the air until it is caught by the receiver? The ball is caught by the receiver when h = 2. h = −4.9t 2+ 3t + 1.75 0.25 = −4.9t 2+ 3t + 1.75 0 = −4.9t 2+ 3t + 1.5 (3) - 4(−4.9)( 1.5)= 38.4 2 Since b 2- 4ac > 0, the equation has two real solutions. ―― __ _ Use the quadratic formula to find the solutions of the quadratic equation. -3 ± √38.4 t = -9.8 -3 ± 6.20 t ≈ -9.8 t ≈ -0.33 or t ≈ 0.94 Disregard the negative solution because there is no negative time in this problem context.The ball was in the air for t ≈ 0.94 second. Elaborate intersect the x-axis at two points. If the discriminant is negative, the equation will have no solutions: the graph will not intersect the x-axis at all. 12. What advantage does using the quadratic formula have over other methods of solving quadratic equations? The quadratic formula works for all quadratic equations. Other methods only work in certain situations. 13. Essential Question Check-In How can you derive the quadratic formula? The quadratic formula is derived by completing the square for the standard form of a quadratic equation. Module 22 A1_MNLESE368187_U8M22L3.indd 1066 1066 © Houghton Mifflin Harcourt Publishing Company · Image Credits: © Corbis Royalty Free 11. How can the discriminant of a quadratic equation be used to determine the number of zeros (x-intercepts) that the graph of the equation will have? If the discriminant is zero, the equation will have one solution: it will intersect the x-axis at one point. If the discriminant is positive, the equation will have two solutions: it will Lesson 3 3/23/14 6:30 AM
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