Quadratic Equations and Applications

Transcription

1 77 Quadratic Equations and Applications Quadratic Equations and Applications Quadratic Equations Recall that a linear equation is one that can be written in the form a b c, where a, b, and c are real numbers, a 0 Quadratic Equation An equation that can be written in the form a b c 0 where a, b, and c are real numbers, with a 0, is a quadratic equation (Why is the restriction a 0 necessary?) A quadratic equation written in the form a b c 0 is in standard form The simplest method of solving a quadratic equation, but one that is not always easily applied, is by factoring This method depends on the following property Zero-Factor Property If ab 0, then a 0 or b 0 or both EXAMPLE Solve Standard form Factor By the zero-factor property, the product 3 3 can equal 0 only if 3 0 or or 3 3 or 3 The solve feature gives the two solutions of the equation in Eample Notice that the guess 0 yields the solution 3, while the guess 5 yields the solution 3 Compare to Eample Check by first substituting and then 3 in the original equation The solution 3 set is, 3 3 A quadratic equation of the form k, k 0, can be solved by factoring k k 0 k k 0 k 0 k or or k 0 k This proves the square root property for solving equations

2 38 CHAPTER 7 The Basic Concepts of Algebra Square Root Property If k 0, then the solutions of k are k If k 0, the equation k has two real solutions If k 0, there is only one solution, 0 If k 0, there are no real solutions (However, in this case, there are imaginary solutions Imaginary numbers are discussed briefly in the Etension on comple numbers at the end of Chapter 6) Completing the square, used in deriving the quadratic formula, has important applications in algebra To transform the epression k into the square of a binomial, we add to it the square of half the coefficient of ; that is, k k We then get For eample, to make 6 the square of a binomial, we add 9, since 9 6 This results in the trinomial 6 9, which is equal to 3 The Greeks had a method of completing the square geometrically For eample, to complete the square for 6, begin with a square of side Add three rectangles of width and length to the right side and the bottom Each rectangle has area or, so the total area of the figure is now 6 To fill in the corner (that is, complete the square ), we add 9 -by- squares as shown k k k EXAMPLE Use the square root property to solve each quadratic equation for real solutions (a) 5 Since 5 5, the solution set is 5, 5, which may be abbreviated 5 (b) r 8 r 8 Square root property r 9 r 9 Product rule for square roots r The solution set is 3 (c) z 3 Since 3 0, there are no real roots, and the solution set is 0 (d) Use a generalization of the square root property, working as follows The solution set is 3 The Quadratic Formula 3 3 By using a procedure called completing the square (see the margin note) we can derive one of the most important formulas in algebra, the quadratic formula We begin with the standard quadratic equation a b c 0, a 0 The new, larger square has sides of length 3 and area b a c a 0 b a c a Divide by a Add c a

3 77 Quadratic Equations and Applications 383 The polynomial on the left will be the square of a binomial if we add to both sides of the equation b a b b b a a a c a a b b ac a b a b ac a b a b ac a b a b ac a b b ac a Factor on the left; combine terms on the right Square root property Quotient rule for square roots Subtract b a Combine terms For centuries mathematicians wrestled with finding a formula that could solve cubic (thirddegree) equations A story from siteenth-century Italy concerns two main characters, Girolamo Cardano and Niccolo Tartaglia In those days, mathematicians often participated in contests Tartaglia had developed a method of solving a cubic equation of the form 3 m n and had used it in one of these contests Cardano begged to know Tartaglia s method and after he was told was sworn to secrecy Nonetheless, Cardano published Tartaglia s method in his 55 work Ars Magna (although he did give Tartaglia credit) The formula for finding one real solution of the above equation is 3 3 n n m 33 n n m 33 Try solving for one solution of the equation using this formula (The solution given by the formula is ) Quadratic Formula The solutions of a b c 0, a 0, are b b ac a Notice that the fraction bar in the quadratic formula etends under the b term in the numerator EXAMPLE 3 Solve 0 Here a, b, and c Substitute these values into the quadratic formula to obtain b b ac a 6 8 a, b, c Factor out a in the numerator Lowest terms The solution set is,, abbreviated

4 38 CHAPTER 7 The Basic Concepts of Algebra A Radical Departure from the Other Methods of Evaluating the Golden Ratio Recall from a previous chapter that the golden ratio is found in numerous places in mathematics, art, and nature In a margin note there, we showed that is equal to the golden ratio, 5 Now consider this nested radical: Let represent this radical Because it appears within itself, we can write Square both sides 0 Write in standard form Using the quadratic formula, with a, b, and c, it can be shown that the positive solution of this equation, and thus the value of the nested radical is (you guessed it!) the golden ratio West 00 FIGURE 3 North Intersection EXAMPLE Solve To find the values of a, b, and c, first rewrite the equation in standard form as 0 Then a, b, and c By the quadratic formula, The solution set is Applications 3 When solving applied problems that lead to quadratic equations, we might get a solution that does not satisfy the physical constraints of the problem For eample, if represents a width and the two solutions of the quadratic equation are 9 and, the value 9 must be rejected, since a width must be a positive number EXAMPLE 5 Two cars left an intersection at the same time, one heading due north, and the other due west Some time later, they were eactly 00 miles apart The car headed north had gone 0 miles farther than the car headed west How far had each car traveled? Step : Read the problem carefully Step : Assign a variable Let be the distance traveled by the car headed west Then 0 is the distance traveled by the car headed north See Figure 3 The cars are 00 miles apart, so the hypotenuse of the right triangle equals 00 Step 3: Write an equation Use the Pythagorean theorem Step : c a b 00 0 Solve 0, Use the quadratic formula to find , or 80 Square the binomial Get 0 on one side Factor out the common factor Divide both sides by a, b 0, c 800 Use a calculator

5 77 Quadratic Equations and Applications 385 Step 5: State the answer Since distance cannot be negative, discard the negative solution The required distances are 60 miles and miles Step 6: Check Since , the answer is correct EXAMPLE 6 If a rock on Earth is thrown upward from a -foot building with an initial velocity of feet per second, its position (in feet above the ground) is given by s 6t t, where t is time in seconds after it was thrown When does it hit the ground? When the rock hits the ground, its distance above the ground is 0 Find t when s is 0 by solving the equation 0 6t t 0 t 7t 9 t t t 8 or t Let s 0 Divide both sides by 6 Quadratic formula Use a calculator Since time cannot be negative, discard the negative solution The rock will hit the ground about 8 seconds after it is thrown