Quadratic Equations

10 Key excerpts on "Quadratic Equations"

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  Technical Mathematics with Calculus

  • Michael A. Calter, Paul A. Calter, Paul Wraight, Sarah White(Authors)
  • 2016(Publication Date)
  • Wiley

    (Publisher)

  Solving quadratics has been enhanced by the evo- lution of today’s scientific calculators. 14–1 Solving Quadratics by Factoring Terminology A polynomial equation of second degree is called a quadratic equation. It is common practice to refer to it simply as a quadratic. Recall that in a polynomial, all powers of x are positive integers. 14 Quadratic Equations 303 Section 14–1 ◆ Solving Quadratics by Factoring ◆◆◆ Example 1: The following equations are Quadratic Equations: (a) 4x 2 - 5x + 2 = 0 (c) 9x 2 - 5x = 0 (b) x 2 = 58 (d) 2x 2 - 7 = 0 Equation (a) is called a complete quadratic; (b) and (d), which have no x terms, are called pure quadratics; and (c), which has no constant term, is called an incomplete quadratic. A quadratic is in general form when it is written in the following form, where a, b, and c are constants: General Form of a Quadratic ax 2 + bx + c = 0 99 ◆◆◆ Example 2: Write the following quadratic equation in general form, then identify the values of a, b, and c, as in Eq. 99, above. (Manipulate the equation.) x x 7 4 5 3 2 - = Solution: Since fractions are generally unpopular when manipulating equations, let’s move the 3 from the fraction x 5 3 2 by multiplying both sides by 3. This allows us to get rid of the 3 in the denominator. Step 1: x x (3) (7 4 ) 5 3 (3) 2 × - = × Let’s expand the left side of the equation. Step 2: 3 . (7 - 4x) = 5x 2 Your equation should look like this so far. Step 3: 21 - 12x = 5x 2 Now move all terms to the left side of the equation in descending order, starting with the x 2 term. Be sure place a zero on the right side of the equation. Step 4: -5x 2 - 12x + 21 = 0 One of the math rules for quadratics is to make sure that the first term is positive, so we can multiply the whole equation by -1. Step 5: (-1) × (-5x 2 - 12x + 21 = 0) We now have the original equation in general form. 5x 2 + 12x - 21 = 0 The equation is now in general form, with a = 5, b = 12, and c = -21.

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  Intermediate Algebra

  • Mark D. Turner, Charles P. McKeague(Authors)
  • 2016(Publication Date)
  • XYZ Textbooks

    (Publisher)

  In this section, we will explore further these types of functions and their graphs. Up to now in this chapter, we have been studying Quadratic Equations. If we take a quadratic equation in standard form and use it instead as the formula for a function, the result is called a quadratic function . Here is a formal definition: Any function that can be written in the form f ( x ) = a x 2 + bx + c where a , b , and c are constants with a ≠ 0, is called a quadratic function . We refer to this form as standard form . quadratic function DEFINITION The Basic Quadratic Function The simplest of all quadratic functions results when we let a = 1 and b = c = 0. We will refer to f ( x ) = x 2 as the basic quadratic function . Table 1 gives some ordered pairs for this function, and the corresponding graph is shown in Figure 1. This graph is an example of a parabola . From the graph we can see that the domain of this function is the set of all real numbers and the range is { y | y ≥ 0}. The point (0, 0) where the parabola changes direction (the lowest point on the parabola) is called the vertex . FIGURE 1 –5 –4 –3 –2 –1 1 2 3 4 5 –5 –4 –3 –2 –1 1 2 3 4 5 Vertex y 5 x 2 2 3 -5 -4 -3 -2 -1 1 2 3 4 5 -1 1 2 3 5 6 7 8 9 x y Vertex f ( x ) = x 2 Axis of Symmetry Axis of Symmetry 4 x f ( x ) = x 2 − 3 9 − 2 4 − 1 1 0 0 1 1 2 4 3 9 TABLE 1 608 CHAPTER 8 Quadratic Equations and Functions We also observe that the graph is symmetric about the y -axis, meaning the right half of the graph is a mirror image of the left half. For this reason, the vertical line x = 0 passing through the vertex is called the axis of symmetry . All quadratic functions have a graph that is a parabola and a domain that is the set of all real numbers. However, the shape, direction, and position of the parabola can vary as we will see in the following segment. Transformations Let's consider quadratic functions of the form f ( x ) = a x 2 .

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  Elementary and Intermediate Algebra

  • Mark D. Turner, Charles P. McKeague(Authors)
  • 2016(Publication Date)
  • XYZ Textbooks

    (Publisher)

  We refer to this form as standard form . quadratic function DEFINITION The Basic Quadratic Function The simplest of all quadratic functions results when we let a = 1 and b = c = 0. We will refer to f ( x ) = x 2 as the basic quadratic function . Table 1 gives some ordered pairs for this function, and the corresponding graph is shown in Figure 1. This graph is an example of a parabola . FIGURE 1 –5 –4 –3 –2 –1 1 2 3 4 5 –5 –4 –3 –2 –1 1 2 3 4 5 Vertex y 5 x 2 2 3 -5 -4 -3 -2 -1 1 2 3 4 5 -1 1 2 3 5 6 7 8 9 x y Vertex f ( x ) = x 2 Axis of Symmetry Axis of Symmetry 4 x f ( x ) = x 2 − 3 9 − 2 4 − 1 1 0 0 1 1 2 4 3 9 TABLE 1 820 Chapter 11 Quadratic Equations and Functions From the graph we can see that the domain of this function is the set of all real numbers and the range is { y | y ≥ 0}. The point (0, 0) where the parabola changes direction (the lowest point on the parabola) is called the vertex . We also observe that the graph is symmetric about the y -axis, meaning the right half of the graph is a mirror image of the left half. For this reason, the vertical line x = 0 passing through the vertex is called the axis of symmetry . All quadratic functions have a graph that is a parabola and a domain that is the set of all real numbers. However, the shape, direction, and position of the parabola can vary as we will see in the following segment. Transformations Let's consider quadratic functions of the form f ( x ) = a x 2 . In this case, all we are doing is taking each y -value (output) from the basic quadratic function y = x 2 and multiplying it by a factor of a . As a result, we can change the shape and direction of the basic parabola. We illustrate how this is done in the next two examples. Graph: f ( x ) = 2 x 2 . SOLUTION Because a = 2, we take each y -coordinate from the basic parabola y = x 2 and double it. Table 2 shows how this is done for several values of x .

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  Technical Mathematics with Calculus

  • Paul A. Calter, Michael A. Calter(Authors)
  • 2011(Publication Date)
  • Wiley

    (Publisher)

  363 12 Quadratic Equations ◆◆◆ OBJECTIVES ◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆ When you have completed this chapter you should be able to • Solve quadratics using a calculator’s equation solver. • Solve quadratics using a calculator that can do symbolic algebra. • Solve quadratics by graphing, either manually or by calculator. • Solve quadratics by the quadratic formula. • Apply quadratics to a variety of applications. ◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆ So far we have only solved first-degree (linear) equations, as well as sets of linear equations. Now we move on to equations of second degree, or Quadratic Equations. We have already solved equations using calculators that have a built-in equation solver, and using calculators that can do symbolic processing. Also, in our chapter on graphing, we learned how to graphically find the approximate solution to any equation. We will start by applying those methods to Quadratic Equations. The meth- ods are no different here, except we must look for two solutions instead of one. We will also show the more traditional manual method, the use of the quadratic formula. As usual, we will follow the mathematics with numerous applications. Take, for example, a simple falling-body problem, Fig.12–1: “If an object is thrown downward with a speed of 15.5 ft/s, how long will it take to fall 125 ft?” If we sub- stitute and into the equation for a freely falling body we get This is called a quadratic equation. How shall we solve it for t? We will show how in this chapter. 125  15.5t  16.1t 2 s  v 0 t  1 2 at 2 a  32.2 ft/s 2 v 0  15.5 ft/s, s  125 ft, FIGURE 12–1 12–1 Solving a Quadratic Equation Graphically and by Calculator Recall that a polynomial equation is one in which all powers of x are positive inte- gers. A quadratic equation is a polynomial equation of second degree. That is, the highest power of x in the equation is 2.

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  Intermediate Algebra

  Concepts with Applications

  • Charles P. McKeague(Author)
  • 2013(Publication Date)
  • XYZ Textbooks

    (Publisher)

  ©iStockphoto.com/fredrocko DEFINITION discriminant The expression under the radical in the quadratic formula is called the discriminant: discriminant = D = b 2 − 4ac OBJECTIVES KEY WORDS Chapter 7 Quadratic Equations and Functions 608 The following table gives the relationship between the discriminant and the type of solutions to the equation. For the equation ax 2 + bx + c = 0 where a, b, and c are integers and a ≠ 0: In the second and third cases, when the discriminant is 0 or a positive perfect square, the solutions are rational numbers. The Quadratic Equations in these two cases are the ones that can be factored. EXAMPLE 1 For each equation, give the number and kind of solutions. a. x 2 − 3x − 40 = 0 Solution Using a = 1, b = −3, and c = −40 in b 2 − 4ac, we have (−3) 2 − 4(1)(−40) = 9 + 160 = 169. The discriminant is a perfect square. The equation therefore has two rational solutions. b. 2x 2 − 3x + 4 = 0 Solution Using a = 2, b = −3, and c = 4, we have b 2 − 4ac = (−3) 2 − 4(2)(4) = 9 − 32 = −23 The discriminant is negative, implying the equation has two complex solutions that contain i. c. 4x 2 − 12x + 9 = 0 Solution Using a = 4, b = −12, and c = 9, the discriminant is b 2 − 4ac = (−12) 2 − 4(4)(9) = 144 − 144 = 0 Because the discriminant is 0, the equation will have one rational solution. d. x 2 + 6x = 8 Solution We must first put the equation in standard form by adding −8 to each side. If we do so, the resulting equation is x 2 + 6x − 8 = 0 Now we identify a, b, and c as 1, 6, and −8, respectively. b 2 − 4ac = 6 2 − 4(1)(−8) = 36 + 32 = 68 The discriminant is a positive number, but not a perfect square. The equation will therefore have two irrational solutions. B Finding an Unknown Constant EXAMPLE 2 Find an appropriate k so that the equation 4x 2 − kx = −9 has exactly one rational solution. Solution We begin by writing the equation in standard form.

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  Elementary Algebra 2e

  • Lynn Marecek, MaryAnne Anthony-Smith, Andrea Honeycutt Mathis(Authors)
  • 2020(Publication Date)
  • Openstax

    (Publisher)

  Then, we do all the math to simplify the expression. The result gives the solution(s) to the quadratic equation. EXAMPLE 10.28 HOW TO SOLVE A QUADRATIC EQUATION USING THE QUADRATIC FORMULA Solve 2x 2 + 9x − 5 = 0 by using the Quadratic Formula. 1184 Chapter 10 Quadratic Equations This OpenStax book is available for free at http://cnx.org/content/col31130/1.4 Solution TRY IT : : 10.55 Solve 3y 2 − 5y + 2 = 0 by using the Quadratic Formula. TRY IT : : 10.56 Solve 4z 2 + 2z − 6 = 0 by using the Quadratic Formula. Chapter 10 Quadratic Equations 1185 If you say the formula as you write it in each problem, you’ll have it memorized in no time. And remember, the Quadratic Formula is an equation. Be sure you start with ‘ x = ’. EXAMPLE 10.29 Solve x 2 − 6x + 5 = 0 by using the Quadratic Formula. Solution This equation is in standard form. Identify the a, b, c values. Write the Quadratic Formula. Then substitute in the values of a, b, c. Simplify. Rewrite to show two solutions. Simplify. Check. TRY IT : : 10.57 Solve a 2 − 2a − 15 = 0 by using the Quadratic Formula. TRY IT : : 10.58 Solve b 2 + 10b + 24 = 0 by using the Quadratic Formula. When we solved Quadratic Equations by using the Square Root Property, we sometimes got answers that had radicals. That can happen, too, when using the Quadratic Formula. If we get a radical as a solution, the final answer must have the radical in its simplified form. HOW TO : : SOLVE A QUADRATIC EQUATION USING THE QUADRATIC FORMULA. Write the Quadratic Formula in standard form. Identify the a , b , and c values. Write the Quadratic Formula. Then substitute in the values of a , b , and c. Simplify. Check the solutions. Step 1. Step 2. Step 3. Step 4. 1186 Chapter 10 Quadratic Equations This OpenStax book is available for free at http://cnx.org/content/col31130/1.4 EXAMPLE 10.30 Solve 4y 2 − 5y − 3 = 0 by using the Quadratic Formula.

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  Algebra

  Form and Function

  • William G. McCallum, Eric Connally, Deborah Hughes-Hallett(Authors)
  • 2015(Publication Date)
  • Wiley

    (Publisher)

  In the previous section we looked at two ways of expressing a quadratic: in factored form and in vertex form. Each way leads to a method for solving the corresponding quadratic equation. Solving Quadratic Equations by Factoring In Example 1 on page 124 we considered the quadratic function which gives the revenue of selling T-shirts at price  dollars, () = (1000 − 20). Since this function is already given as a product of linear factors, we can find its zeros by finding where the factors are zero. In general, we have: The Zero-Factor Principle If  ⋅  = 0 then either  = 0 or  = 0 (or both). According to the zero-factor principle, the only possibilities for  ⏟ ⏟ ⏟  (1000 − 20) ⏟⏞⏞⏞⏞⏞ ⏟⏞⏞⏞⏞⏞ ⏟  = 0 are  = 0 or 1000 − 20 = 0. This gives the solutions  = 0 and  = 50. If we can factor a quadratic equation written in standard form easily, then we can use the zero- factor principle to find the solutions. Example 1 Solve the Quadratic Equations using the zero-factor principle and factoring. (a)  2 − 4 + 3 = 0 (b) 3 2 + 3 = 6 3.4 Quadratic Equations 117 Solution (a) We have  2 − 4 + 3 = ( − 1)( − 3) = 0, so we want to solve the equation ( − 1) ⏟ ⏟ ⏟  ( − 3) ⏟ ⏟ ⏟  = 0. If  is a solution, then either  = 0, which implies  − 1 = 0  = 1, or  = 0, which implies  − 3 = 0  = 3. (b) In order to use the zero-factor principle, we must first write the quadratic equation in the standard form by subtracting 6 from both sides. This gives 3 2 + 3 − 6 = 0  2 +  − 2 = 0 divide both sides by 3 ( + 2)( − 1) = 0. This implies  = −2 and  = 1 are the solutions. In general, the zero-factor principle tells us that for a quadratic equation in the form ( − )( − ) = 0, where , ,  are constants,  ≠ 0, the only solutions are  =  and  = . Solving Quadratic Equations with Perfect Squares Factoring is a useful technique for solving Quadratic Equations, but sometimes factoring may be difficult (or impossible).

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  Beginning Algebra

  Connecting Concepts through Applications

  • Mark Clark, Cynthia Anfinson(Authors)
  • 2018(Publication Date)
  • Cengage Learning EMEA

    (Publisher)

  These two solutions can be written as x 5 6 2 . Solving the quadratic equation x 2 5 4 using the square root property yields x 2 5 4 The only variable term, x 2 , is isolated. x 5 6 ! 4 Use the square root property. x 5 6 2 This means x 5 2 or x 5 2 2 . Check: Check both solutions in the original problem. x 5 2 x 5 2 2 ( 2 ) 2 0 4 ( 2 2 ) 2 0 4 4 5 4 Both answers check. 4 5 4 This new notation we are using, x 5 6 2 , means that x 5 2 or x 5 2 2 . It is a short-hand way to write both solutions compactly. When a negative number is squared, the answer is positive. Connecting the Concepts What is a quadratic equation? In Section 6.4, a quadratic equation was defined as an equation that can be put in the form ax 2 1 bx 1 c 5 0 , where a ? 0 . We call this the general form of a quadratic equation. Copyright 2019 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. S E C T I O N 9 . 2 S o l v i n g Q u a d r a t i c E q u a t i o n s b y U s i n g t h e S q u a r e R o o t P r o p e r t y 765 Steps to Solving Quadratic Equations Using the Square Root Property 1. Determine whether the quadratic equation is of the form x 2 5 a . 2. Isolate the squared variable term if needed. 3. Use the square root property. If x 2 5 a and a . 0 then x 5 6 ! a . 4. Solve for the variable. 5. Check both the answers in the original equation. Example 1 The falling rock The height, h , in feet from the ground, of a falling rock t seconds after being dropped from a 40-foot cliff is given by the equation h 5 2 16 t 2 1 40 a. How high above the ground is the rock after falling 0.5 second? b.

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  Elementary Algebra

  • Alan Tussy, R. Gustafson(Authors)
  • 2012(Publication Date)
  • Cengage Learning EMEA

    (Publisher)

  9.3 Solving Quadratic Equations: The Quadratic Formula 699 NOTATION Complete the solution. 11. Solve: 12. Fill in the blanks: To read , we say, “the of , plus or the square root of minus , all .” GUIDED PRACTICE Use the quadratic formula to solve each equation. See Example 1. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. Use the quadratic formula to solve each equation. Approximate the solutions to nearest hundredth. See Example 2. 25. 26. 27. 28. 29. 30. 31. 32. 33. 34. 3 x 2 5 x 4 4 x 2 7 x 2 4 x 2 x 2 7 x 2 3 x 1 5 x 2 3 x 1 3 x 2 x 3 x 2 2 3 x x 2 4 7 x x 2 2 5 x x 2 1 3 x 2 x 2 x 1 0 5 x 2 13 x 6 0 3 x 2 8 x 3 0 3 x 2 5 x 2 0 4 x 2 4 x 3 0 6 x 2 5 x 6 0 2 x 2 3 x 2 0 4 x 2 3 x 1 0 x 2 8 x 15 0 x 2 7 x 12 0 x 2 5 x 4 0 x 2 5 x 6 0 2 a 4 ac b b b 2 b 2 4 ac 2 a x 5 2 or x 5 2 x 7 2 x 5 2 2 x 5 2 25 2 x ( ) 2 ( 5) 2 4(1)( ) 2( ) x b 2 b 2 4 ac 2 a x 2 5 x 6 0 VOCABULARY Fill in the blanks. 1. The general equation is , where . 2. Complete the quadratic formula. CONCEPTS 3. Write each equation in form. a. b. 4. For each quadratic equation, find , , and . a. b. 5. Divide both sides of by 2, and then find , , and . 6. Evaluate each expression. a. b. 7. a. How many terms does the expression have? b. What common factor do the terms have? 8. Simplify each expression. a. b. 9. A student used the quadratic formula to solve an equation and obtained a. How many solutions does the equation have? b. What are they? c. Approximate each to the nearest hundredth. 10. Match each quadratic equation with the best method for solving it. Each answer can be used only once. a. i. Square root method b. ii. Factoring method c. iii. Complete the square d. iv. Quadratic formula x 2 10 x 5 x 2 21 x 2 19 x 2 0 x 2 7 x 8 0 x 3 2 15 2 ( 4) 2 ( 4) 2 4(2)( 9) 2(2) 1 2 45 2(7) 10 15 2 2 ( 1) 2 ( 1) 2 4(2)( 4) 2(2) 2 2 2 2 4(1)( 8) 2(1) c b a 2 x 2 4 x 8 0 8 x 2 x 10 x 2 5 x 6 0 c b a 3 x 2 2 x 1 x 2 2 x 5 ax 2 bx c 0 x a 0 ax 2 bx c 0 STUDY SET SECTION 9.3 Copyright 201 Cengage Learning.

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  Intermediate Algebra

  Connecting Concepts through Applications

  • Mark Clark, Cynthia Anfinson(Authors)
  • 2018(Publication Date)
  • Cengage Learning EMEA

    (Publisher)

  1 x 1 321 x 2 721 x 1 152 5 0 From this Concept Investigation, we see that equations that are factored and set equal to zero can be solved by setting each factor equal to zero one at a time. By factoring a polynomial first, we then can solve the simpler equations. It is very important to note that we can use the factored form to solve an equation only if it is set equal to zero. If the quadratic equation is not equal to zero, first move everything to one side so that it equals zero and then factor. How does factoring help us solve? CONCEPT INVESTIGATION Steps to Solving Polynomial Equations by Factoring 1. Set the polynomial equal to zero. 2. Factor the polynomial completely. 3. Use the zero factor property to set each of the factors equal to zero and solve. 4. Check the answer(s) in the original equation. Skill Connection AC method of factoring ax 2 1 bx 1 c 1. Factor out any common factors. 2. Multiply a and c together. 3. Find factors of ac that add up to b. 4. Rewrite the middle (bx) term using the factors from step 3. 5. Group and factor out what is in common. To review factoring, see the Factoring Toolbox or Chapter 3. Example 1 Solving polynomial equations by factoring Solve the following by factoring. Check the answers. a. 2x 2 1 27 5 21x b. x 3 1 8x 2 1 15x 5 0 SOLUTION a. Step 1 Set the polynomial equal to zero. 2x 2 1 27 5 21x 2x 2 2 21x 1 27 5 0 Step 2 Factor the polynomial completely. 2x 2 2 3x 2 18x 1 27 5 0 1 2x 2 2 3x 2 1 1 218x 1 272 5 0 x 1 2x 2 32 2 91 2x 2 32 5 0 1 2x 2 321 x 2 92 5 0 (x +2)(x - 5) Quadratic Equations Square Root Property Use when there is a squared term but no first degree term. Isolate the squared term and use a plus or minus symbol to indicate both answers. x 2 5 25 x 5 6 !25 x 5 65 Completing the Square Use if the vertex form is required.

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