1.6-Quadratic Equations

Transcription

1 1.6-Quadratic Equations A quadratic equation is any equation that can be written in the form a + b + c = where a, b, and c are real numbers and a. The following are eamples of quadratic equations = 4 5 = = 6 There are a number of methods to use depending on the nature of the equation. Some equations may be solved by multiple methods while others may only be solved using one method. Solving Quadratic Equations by Factoring Many quadratic equations may be solved by factoring and using the Zero Factor Property. To solve by factoring the quadratic equation must be set equal to zero. Zero Factor Property: If A and B are algebraic epressions, then the equation AB = is equivalent to the compound statement A = or B =. Eample: Solve the equation = by factoring. Solution: Since the equation is already set equal to zero, we may begin by factoring. = ( 5)( + 4) = Now we use the zero product property and set each linear factor equal to zero and solve. ( 5)( + 4) = ( 5) =,( + 4) = 5, 4 The solutions are 5, and -4. These may be verified by checking in the original equation. Eample: Solve the equation 5 3 = by factoring. Solution: Since the equation is already set equal to zero, we may begin by factoring. 5 3 = ( + 1)( 3) = Now we use the zero product property and set each linear factor equal to zero and solve. ( + 1)( 3) = ( + 1) =,( 3) = 1, 3 The solutions are 3, and - 1. These may be verified by checking in the original equation.

2 Eample: Solve the equation ( 3)( + 4) = 3 by factoring. Solution: We must first multiply the factors on the left side of the equation together. Then set the equation equal to zero and factor. ( 3)( + 4) = = = ( + 7)( 6) = Now we use the zero product property and set each linear factor equal to zero and solve. ( + 7)( 6) = ( + 7) =,( 6) = 7, 6 The solutions are -7, and 6. These may be verified by checking in the original equation. The Square Root Property The square root property is based on the inverse relationship between and, that is these two operations will undo or cancel each other out. This means that equations that are in the form a c. We use this fact to solve quadratic Eample: Solve the equation t = 3by using the square root property. Solution: The equation is already in the form a c so apply the square root property. t t = 3 = ± 3 t = ± 4 Eample: Solve the equation t + 18 = by using the square root property. Solution: Rewrite the equation in the form a c then apply the square root property. t + 18 = t = 18 t = ± 18 t = ± 3i

3 Eample: Solve the equation 4 8 = by using the square root property. Solution: Rewrite the equation in the form a c then apply the square root property = = 8 = 7 ± = ± Solving a Quadratic Equation by Completing the Square Completing the square is very useful method for solving quadratic equations. This procedure involves first completing the square and then using the square root property. Eample: Solve + 6 = by completing the square. 7 7 Solution: First, complete the square. + 6 = = 9 ( + 3) = 9 Notice that 9 needed to be added to both sides of the equation. Second, solve by using the square root property. ( + 3) = 9 ( + 3) = ± + 3 = ± 3 3 ± 3, 6 9 What happens if there is already a c term? If this occurs then the c term must be moved to the other side of the equation. Eample: Solve + 1 = by completing the square. Solution: First, move -1 to the right side of the equation, then complete the square. + = = ( + 1) = 13 Notice that 1 needed to be added to both sides of the equation.

4 Second, solve by using the square root property. ( + 1) ( + 1) + 1 = ± 1± = 13 = ± Notice that there are two solutions. This is to be epected since we are solving a quadratic equation. Be sure to check both solutions. If the a term in by a. a + b + c is not 1, then we need to make it a 1 by dividing both sides of the equation Eample: Solve 6 5 = 4 + by completing the square. Solution: First, get all terms on the correct side of the equation and then divide both sides of equation by = = = Second, complete the square = = Now solve by using the square root property = ± = 5 ± Notice that there are two irrational solutions. This will happen very frequently. Always leave the answer in simplified radical form and be sure to check both solutions

5 Solving Quadratic Equations Using the Quadratic Formula When solving a quadratic equation, using the quadratic formula it is important to first write the equation in the form a + b + c =. Once this is accomplished, identify the a, b, and c values. Finally, plug these values into the quadratic formula and simplify completely. Eample: Solve the equation = 18 using the quadratic formula. Solution: Rewriting the equation in the form a + b + c = gives the equation =. Then a = 3, b = 15, and c = -3. The Discriminant b ± 15 ± 15 ± 15 ± 6 15 ± ± b 4ac a (15) (3) 4(3)( 3) The discriminant b 4ac is the part of the quadratic formula underneath the radical sign, that is, it is the radicand of the quadratic formula. That radicand can tell us about the number and type of solutions to a quadratic equation as follows: If b 4ac > then there will be two real solutions to the quadratic equation. If b 4ac = then there will be one real solution to the quadratic equation. If b 4ac < then there will be two imaginary solutions to the quadratic equation. Be careful. The discriminant does NOT tell us the actual solutions, but rather, only the number and type of solutions to a quadratic equation. Eample: Use the discriminant to determine the number and type of solutions to the equation =. Solution: We can determine that a = 5, b =, and c =. Then, b 36 4ac 4(5)() Since the discriminant is 36 < there are two imaginary solutions. Refer to eample 5 for the actual solutions.

6 1.6-Applications The Ballistics Equation: 1 gt + vt s s = +, is used in many applications that involve a self-propelled object moving through the air. In the ballistics equation s represents a position at time t; g is the acceleration due to gravity, v represents initial velocity and s represents initial position. g is either 3 fps or 9.8mps depending on the measurement system. Eample: A juggler tosses a ball into the air with a velocity of 4 ft/sec. from a height of 4 ft. Find how long it takes the ball to return to a height of 4 ft. Solution: This will require the use of the ballistics equation. In this problem g = 3 fps, v = 4, and s = 4. Substituting these values into the given equation gives the following equation. 1 s = gt + vt + s s = 16t + v t + s 4 = 16t Rewrite the equation in standard form and solve. 4 = 16t 16t + 4t + 4 4t = + 4t + 4 8t(t 5) = 8t = t 5 = t = t =.5 The ball will be at 4 feet at time seconds and.5 seconds.

7 Eample: The perimeter of a rectangular flag is 34 in. and the diagonal is 13 in. What are the length and the width of the flag? Solution: The solution to this problem is based on two basic geometric formulas; the perimeter of a rectangle P = L + W and the Pythagorean Theorem A + B = C. Starting with the perimeter formula, let the length of the rectangle. Then the width will be: 34 = L + W 17 = L + W 17 = + W W = 17 We may now substitute all known values into the Pythagorean Theorem where C will represent the diagonal of the rectangle, A will represent the length and B will represent the width. A + B + (17 ) = = C 34 + = = = 169 This equation may be solved by factoring = ( 1)( 5) = 1 =, 5 = 1, 5 Since we defined the length to be, substitute these values into the width formula to determine the width for each value of the length. W = 17 W = 17 W = 17 5 W = 1 W = 17 1 W = 5 Therefore, the length and width are 5 and 1, and it does not matter which is which. Both of these answers check in both the original formulas.

8 Eample: A sky diver steps out of an airplane at 5, ft. Use the ballistics equation to find how long it takes the sky diver to reach 4, ft. Solution: Let g = 3 fps, v =, and s = 5,. Substituting these values into the given equation gives the following equation. s = 16t 4, = 16t + v t + s + 5, Since there is no b term, I will use the square root property to solve for. 4, = 16t 16t t = 6.5 t = ± t = 7.9 = 6.5 It will take the skydiver 7.9 seconds to fall to 4, feet. + 5, Eample: The total cost of producing items is given byc = For what number of items is the average cost per item equal to $5.5? Solution: The average cost may be determined by dividing the total cost by the number of items. Setting this equal to 5.5 gives the following equation = 5.5 Factor and cancel an on the left side of the equation. Since there is no b term, I will use the square root property to solve for = 5.5. = = ± 5 = The negative value may be disregarded, so the only answer is 5.

9 Eample: Casey wants to make an open top bo for packing baked goods by cutting equal squares from each corner of an 11 in by 14 in. piece of cardboard. He figures that the area of the bottom must be 8 sq. in. What size should he cut from each corner? Solution: We will begin by letting equal the length of each corner cut out from the rectangle. Therefore, the length of the bo will be 14- and the width will be 11-. Now, we will need the area of a rectangle formula A=LW. Substitute the know values into the formula to obtain a quadratic equation. I will use the quadratic equation to solve. A = LW 8 = (14 )(11 ) 8 = = = 5 ± ( ) () 4()(37) 5 ± The value.79 may be disregarded since it is too large. The only valid answer then is 1.7

10 Eample: Atmospheric pressure a decreases with altitude above sea level h. The equation models this situation a = 3.89 h 3.48 h + 1. Find a at 18, ft above sea level. The atmospheric pressure at the highest human settlement is.5. What is the altitude of this settlement? Solution: To solve the first part of the problem, substitute the known values into the equation and evaluate to find a. a = 3.89 a = 3.89 a = a =.499 h 3.48 (18,) h (18,) + 1 To solve the second part of the problem, we need to solve the equation for h. a = = This will require use of the quadratic formula ,44.15 h h 3.48 h 3.48 ( 3.48 ( ± ± ,17.99 or ± 3.48 ) h + 1 h + 1 h +.48 = 4(3.89 ) )(.48) Since the highest point on the Earth is only 8, feet above sea level, the only logical answer is 17,17.99 feet above sea level.

11 Eample: The demand equation for a certain product is P=4-.1, where is the number of units sold per week and P is the price in dollars at which each one is sold. The weekly revenue R is given by R = P.What number of units sold produces weekly revenue of $175,? Solution: First, we will create the revenue equation R = P = (4.1) R = 4.1 Net, substitute 175, into the equation for R and set equal to zero. 175, = , = This is the quadratic equation that must be solved for. Any appropriate method may be used, so I will use the quadratic formula. 4 ± (4) 4 ± 9. 5, or 35, 4(.1)(175,) (.1) The weekly revenue will be $175, when either 5, or 35, units are sold. Although this may not at first make sense, when actual market dynamics are considered it should become quite clear.

12 Eample: The amount spent on marketing by Windbreaker Airline over the past year is modeled by the marketing equation d = m 13m + 4 where d is marketing dollars in thousands and m is months. The company realizes that there is a strong relationship between the amount spent on marketing and the revenue generated each month. It is consequently concerned that there are some months where there are no earnings being spent on marketing. If Windbreaker increases its marketing by $5 per month, will this guarantee that it will always be spending some of its earnings on marketing? Solution: It should be obvious that the marketing equation is a quadratic equation. If Windbreaker increases marketing dollars by $5, we obtain the new equation d d = m = m 13m m + 54 Since we only need to know if d = will ever eist, we can use the discriminant to obtain: b 4ac ( 13) 169,16 19,991 4(1)(54) Because the discriminant is negative, this means that the marketing equation has no real solutions which suggest that marketing dollars will never equal zero.