Solving Quadratic Equations

11 Key excerpts on "Solving Quadratic Equations"

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  eBook - ePub

  Years 9 - 10 Maths For Students

  • (Author)
  • 2015(Publication Date)
  • For Dummies

    (Publisher)

  Q uadratic (second-degree) equations are nice to work with because they’re manageable. Finding the solution or deciding whether a solution exists is relatively easy — easy, at least, in the world of mathematics.

  A quadratic equation is a quadratic expression with an equal sign attached. As with linear equations (covered in Chapter 10 ), specific methods or processes, given in detail in this chapter, are employed to successfully solve quadratic equations. The most commonly used technique for solving these equations is factoring, but a quick-and-dirty rule can also be used for one of the special types of quadratic equations. I have to warn you, though, that just because someone puts in some numbers and makes up a quadratic equation, that doesn’t mean there’s necessarily a solution or answer to it. (I show you how to tell if there’s no answer in this chapter.)

  Quadratic equations are important to algebra and many other sciences. Some quadratic equations say that what goes up must come down. Other equations describe the paths that planets and comets take. In all, quadratic equations are fascinating — and just dandy to work with.

  Squaring Up to Quadratics

  A quadratic equation contains a variable term with an exponent of 2 and no variable term with a higher power.

  A quadratic equation has a general form that goes like this: ax2 + bx + c = 0. The constants a, b and c in the equation are real numbers, and a cannot be equal to 0. (If a were 0, you wouldn’t have a quadratic equation anymore.)

  If the equation looks familiar, it means that you’ve read Chapter 9 , which talks about factoring and working with quadratic expressions. Remember: An expression consists of one or more terms but has no equal sign. Adding an equal sign changes the whole picture: Now you have an equation that says something. The equation forms a true statement if the solutions are put in for the variables.

  Here are some examples of quadratic equations and their solutions:

  • 4x2 + 5x − 6 = 0: In this equation, none of the coefficients is 0. The two solutions are x = −2 and

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  Foundations of GMAT Math

  • Manhattan GMAT(Author)
  • 2011(Publication Date)
  • Manhattan Prep Publishing

    (Publisher)

  Chapter 7: Quadratic Equations

  In This Chapter:

    •  Manipulating quadratic expressions and Solving Quadratic Equations Mechanics of Quadratic Equations In high school algebra, you learned a number of skills for dealing with quadratic equations. For the GMAT, you need to relearn those skills.

  Let's define terms first. A quadratic expression contains a squared variable, such as x2 , and no higher power. The word “quadratic” comes from the Latin word for “square.” Here are a few quadratic expressions:

  z2

  y2 + y – 6

  x2 + 8x + 16

  w2 – 9

  A quadratic expression can also be disguised. You might not see the squared exponent on the variable explicitly. Here are some disguised quadratic expressions.

  z × z

  (y + 3)(y – 2)

  (x + 4)2

  (w – 3)(w + 3)

  If you multiply these expressions out—that is, if you distribute them—then you see the exponents on the variables. Note that the second list corresponds to the first list exactly.

  A quadratic equation contains a quadratic expression and an equals sign.

  Quadratic expression = something else

  A quadratic equation usually has two solutions. That is, in most cases, two different values of the variable each make the equation true. Solving a quadratic equation means finding those values.

  Before you can solve quadratic equations, you have to be able to distribute and factor quadratic expressions.

  Distribute (a + b)(x + y) Use FOIL

  Recall that distributing means applying multiplication across a sum.

  =	5 × 3	+	5 × 4
  equals	five times three	plus	five times four.

  You can omit the multiplication sign next to parentheses. Also, the order of the product doesn't matter, and subtraction works the same way as addition. Here are more examples:

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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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  Technical Mathematics with Calculus

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

    (Publisher)

  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. It is common practice to refer to a quadratic equation simply as a quadratic. ◆◆◆ Example 1: The following equations are quadratic equations: (a) (b) (c) (d) (e) ◆◆◆ A quadratic function is one whose highest-degree term is of second degree. ◆◆◆ Example 2: The following functions are quadratic functions: (a) (b) (c) (d) ◆◆◆ Some quadratic equations have a term missing. A quadratic that has no x term is called a pure quadratic; one that has no constant term is called an incomplete quadratic. ◆◆◆ Example 3: (a) is a pure quadratic. (b) is an incomplete quadratic. ◆◆◆ Solving a Quadratic Graphically We will show several ways to solve a quadratic; first graphically and by calculator, and, in the next section, by formula. ■ Exploration: In our chapter on graphing we plotted the quadratic function getting a curve that we called a parabola. Try this. Either graph this function again or look back at our earlier graph. Does the curve intercept the x axis, and if so, how many times? Can you imagine a parabola that has more x-intercepts? Zoom out far enough to convince yourself that the curve will not turn and re-cross the x axis at some other place. Can you imagine a parabola that has fewer x-intercepts? Using , or zero, find the value of x at an intercept and substitute it into the given function. Then state the significance of an intercept. ■ Your exploration may have shown you that a quadratic function in x can have 0, 1, or 2 x-intercepts, but no more than two. Also, the value of x at an intercept, when substituted back into the function, makes that function equal to zero.

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  Intermediate Algebra

  Connecting Concepts through Applications

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

    (Publisher)

  Many models that we encounter are not easily factored. We now generate a formula that will solve any quadratic equation. Starting with a quadratic equation in standard form, we can solve for x by completing the square. The only requirement is that a ? 0, which guarantees the equation is quadratic and division by 0 will not occur. ax 2 1 bx 1 c 5 0 ax 2 1 bx 5 2c x 2 1 b a x 5 2c a x 2 1 b a x 1 b 2 4a 2 5 2c a 1 b 2 4a 2 a x 1 b 2a b 2 5 b 2 2 4ac 4a 2 x 1 b 2a 5 6 Å b 2 2 4ac 4a 2 x 1 b 2a 5 6 !b 2 2 4ac 2a x 5 2b 2a 6 !b 2 2 4ac 2a x 5 2b 6 !b 2 2 4ac 2a Set the standard quadratic expression equal to zero. Move the constant to the other side of the equal sign. Divide by a to make the leading coefficient 1. Note: a ≠ 0 so that division by a is defined. Add a constant that completes the square (half of the coefficient of x, squared). Factor and simplify. Use the square root property. Simplify the denominator of the radical. Isolate x by subtracting b 2a from both sides. Solving Quadratic Equations by Using the Quadratic Formula LEARNING OBJECTIVES Solve a quadratic equation using the quadratic formula. Solve systems of equations involving quadratic functions. 4.6 What’s That Mean? Quadratic Formula Be careful with the difference between the following: • Quadratic formula: a formula that is used to solve quadratic equations x 5 2b 6 !b 2 2 4ac 2a • Quadratic equation: the equation that is being solved 2x 2 1 3x 2 5 5 20 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.

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  Making Sense of Mathematics for Teaching High School

  Understanding How to Use Functions

  • Edward C. Nolan, Juli K. Dixon, Farhsid Safi, Erhan Selcuk Haciomeroglu(Authors)
  • 2016(Publication Date)
  • Solution Tree Press

    (Publisher)

  Throughout high school, students extend their interaction with linear equations to include representing and solving systems of equations graphically. Following explorations from grade 8 when students analyzed and solved linear equations and solved systems of linear equations, high school students make sense of the solution of systems of linear equations and connect multiple representations of systems of equations to one another. In addition, they prove that the method of substitution or elimination produces a system with the same solutions.

  Students explore quadratic functions in the standard, factored, and vertex forms. They derive the vertex form and the quadratic formula by completing the square so that they can apply these forms to represent, understand, and solve real-world and mathematical problems. Students will further explore and apply quadratic functions and other functions in later high school work.

  The Mathematics

  The ability to represent both real-world and mathematical situations in many different ways is important to solve problems. Students develop flexibility in the ways they mathematize situations by exploring multiple representations of functions and by justifying the procedures they use. Learning to create and use these representations is important in solving problems efficiently. High school students must be able to model contexts through visualization, explore different forms of linear functions, and explore different forms of quadratic equations.

  Modeling Contexts Through Visualization

  Students learn to model contexts with mathematics in elementary school and the middle grades. This often occurs in ways that are different than the ways that many teachers learned to make sense of these situations. Consider the middle school bridge task provided in figure 2.5 . How would you solve it?

  Figure 2.5: Bridge task.

  Often teachers reason in ways that mask some aspects of a mathematical relationship. For example, you might make a table to solve this problem similar to the one provided in table 2.1 . How could you use this table to find the number of beams in a bridge of length n

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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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  The Mathematics That Every Secondary School Math Teacher Needs to Know

  • Alan Sultan, Alice F. Artzt(Authors)
  • 2017(Publication Date)
  • Routledge

    (Publisher)

  Let us now illustrate a typical secondary school problem where a quadratic equation is solved by completing the square. Notice that this method requires that a, the coefficient of x 2, be equal to 1. Example 3.15 Using the method of completing the square, solve the equation x 2 + 6 x + 1 = 0. Solution : We subtract 1 from each side of the equation to get x 2 + 6 x = −1. We complete the square on the left side by adding 9. Of course to keep the equation balanced, we need to do the same to the right hand side. Our equation becomes: x 2 + 6 x + 9 = −1 + 9. This is the same as (x + 3) 2 = 8. Thinking of (x + 3) as y, this tells us we have y 2 = 8 and hence y = ± 8. Replacing y by x + 3 we have, x + 3 = ± 8. So x = - 3 ± 8. Example 3.16 Use the method of completing the square to solve the equation 3 x 2 + 4 x − 2 = 0. Solution : We add 2 to both sides to get 3 x 2 + 4 x = 2. To use the method of completing the square, we need the coefficient of x 2 to be 1. So we divide the equation by 3 to get x 2 + 4 3 x = 2 3. We add [ 1 2 (4 3) ] 2 = 4 9 to both sides of the equation to get x 2 + 4 3 x + 4 9 = 2 3 + 4 9 which just becomes x + 2 3 2 = 10 9. From this we get that (x + 2 3) = ± 10 9, and so, x = - 2 3 ± 10 9. It is exactly in this way that we derive the quadratic formula. Here it is for completeness. Example 3.17 Derive the quadratic formula. Solution : We start with the equation ax 2 + bx + c = 0 where a > 0. We then subtract c from both sides to get a x 2 + b x = - c. Since we need the coefficient of x 2 to be 1, we divide both sides by a to get x 2 + b a x = - c a. We complete the square on the left side by adding (1 2 ⋅ b a) 2, or just b 2 4 a 2. We get x 2 + b a x + b 2 4 a 2 = - c a + b 2 4 a 2. (3.9) Now the left side of equation (3.9) is a perfect square,. the square of (x + b 2 a). Thus, we have (x + b 2 a) 2 = - c a + b 2 4 a 2 which can be rewritten as (x + b 2 a) 2 = - 4 a c 4 a 2 + b 2 4 a 2. Combining the two fractions on the right we

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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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  Elementary and Intermediate Algebra

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

    (Publisher)

  What is interesting about this last example is that it has rational solutions, meaning it could have been solved by factoring. But looking back at the equation, factoring does not seem like a reasonable method of solution because the coefficients are either very large or very small. So, there are times when using the quadratic formula is a faster method of solution, even though the equation you are solving is factorable. 848 Chapter 11 Quadratic Equations and Functions More About Example 5 To visualize the functions in Example 5, we set up our calculator this way: Y 1 = 35X − .1X 2 Revenue function Y 2 = 8X + 500 Cost function Y 3 = Y 1 − Y 2 Profit function Window: X from 0 to 350, Y from 0 to 3,500 Graphing these functions produces graphs similar to the ones shown in Figure 2. The lowest graph is the graph of the profit function. Using the Zoom and Trace features on the lowest graph at Y 3 = 1,200 produces two corresponding values of X, 170 and 100, which match the results in Example 5. 0 350 0 3500 Y 1 Y 2 Y 3 FIGURE 2 Graphing Calculators USING TECHNOLOGY Optimization The term optimization refers to the process of finding an optimal solution to some problem. In many cases, the optimal solution will be the maximum value or minimum value for some quantity. For example, a business would be interested in finding the maximum profit or minimum cost. If the quantity in question is being modeled by a quadratic function, then the maximum or minimum value of the function (the optimal value) will correspond to the vertex. For the water balloon described in Example 3, find the maximum height reached by the water balloon, and determine when it reaches this height. SOLUTION From Example 3, the function giving the height of the water balloon after t seconds is h ( t ) = − 16 t 2 + 40 t + 12 The graph of h ( t ) is a parabola that opens downward, so the highest point on the graph is the vertex.

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  Elementary Algebra

  • Mark D. Turner, Charles P. McKeague(Authors)
  • 2017(Publication Date)
  • XYZ Textbooks

    (Publisher)

  My life was actually going in that direction so I decided to see where that would take me. It was a good decision. It is a good idea to work hard toward your goals, but it is also a good idea to take inventory every now and then to be sure you are headed in the direction that is best for you. I wish you good luck with the rest of your college years, and with whatever you decide you want to do as a career. Pat McKeague 619 9.5 Linear Equations Learning Objectives Learning Objectives In this section, we will learn how to: 1. Write square roots of negative numbers in terms of i . 2. Solve quadratic equations having complex solutions. 9.5 Introduction The quadratic formula tells us that the solutions to equations of the form ax 2 + bx + c = 0 are always: x = − b ± √ — b 2 − 4 ac ______________ 2 a The part of the quadratic formula under the radical sign is called the discriminant : Discriminant = b 2 − 4 ac When the discriminant is negative, we have to deal with the square root of a negative number. We handle square roots of negative numbers by using the definition i = √ — − 1. To illustrate, suppose we want to simplify an expression that contains √ — − 9, which is not a real number. We begin by writing √ — − 9 as √ — 9( − 1) . Then, we write this expression as the product of two separate radicals: √ — 9 ∙ √ — − 1 . Applying the definition i = √ — − 1 to this last expression, we have √ — 9 ∙ √ — − 1 = 3 i As you may recall from the previous section, the number 3 i is called a complex number. Here are some further examples. Write the following radicals as complex numbers: SOLUTION a. √ — − 4 = √ — 4( − 1) = √ — 4 ∙ √ — − 1 = 2 i b. √ — − 36 = √ — 36( − 1) = √ — 36 ∙ √ — − 1 = 6 i c. √ — − 7 = √ — 7( − 1) = √ — 7 ∙ √ — − 1 = i √ — 7 d. √ — − 75 = √ — 75( − 1) = √ — 75 ∙ √ — − 1 = 5 i √ — 3 In parts (c) and (d) of Example 1, we wrote i before the radical because it is less confusing that way.

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