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Algebra and Trigonometry

2.5 Quadratic Equations

Algebra and Trigonometry2.5 Quadratic Equations
  1. Preface
  2. 1 Prerequisites
    1. Introduction to Prerequisites
    2. 1.1 Real Numbers: Algebra Essentials
    3. 1.2 Exponents and Scientific Notation
    4. 1.3 Radicals and Rational Exponents
    5. 1.4 Polynomials
    6. 1.5 Factoring Polynomials
    7. 1.6 Rational Expressions
    8. Key Terms
    9. Key Equations
    10. Key Concepts
    11. Review Exercises
    12. Practice Test
  3. 2 Equations and Inequalities
    1. Introduction to Equations and Inequalities
    2. 2.1 The Rectangular Coordinate Systems and Graphs
    3. 2.2 Linear Equations in One Variable
    4. 2.3 Models and Applications
    5. 2.4 Complex Numbers
    6. 2.5 Quadratic Equations
    7. 2.6 Other Types of Equations
    8. 2.7 Linear Inequalities and Absolute Value Inequalities
    9. Key Terms
    10. Key Equations
    11. Key Concepts
    12. Review Exercises
    13. Practice Test
  4. 3 Functions
    1. Introduction to Functions
    2. 3.1 Functions and Function Notation
    3. 3.2 Domain and Range
    4. 3.3 Rates of Change and Behavior of Graphs
    5. 3.4 Composition of Functions
    6. 3.5 Transformation of Functions
    7. 3.6 Absolute Value Functions
    8. 3.7 Inverse Functions
    9. Key Terms
    10. Key Equations
    11. Key Concepts
    12. Review Exercises
    13. Practice Test
  5. 4 Linear Functions
    1. Introduction to Linear Functions
    2. 4.1 Linear Functions
    3. 4.2 Modeling with Linear Functions
    4. 4.3 Fitting Linear Models to Data
    5. Key Terms
    6. Key Concepts
    7. Review Exercises
    8. Practice Test
  6. 5 Polynomial and Rational Functions
    1. Introduction to Polynomial and Rational Functions
    2. 5.1 Quadratic Functions
    3. 5.2 Power Functions and Polynomial Functions
    4. 5.3 Graphs of Polynomial Functions
    5. 5.4 Dividing Polynomials
    6. 5.5 Zeros of Polynomial Functions
    7. 5.6 Rational Functions
    8. 5.7 Inverses and Radical Functions
    9. 5.8 Modeling Using Variation
    10. Key Terms
    11. Key Equations
    12. Key Concepts
    13. Review Exercises
    14. Practice Test
  7. 6 Exponential and Logarithmic Functions
    1. Introduction to Exponential and Logarithmic Functions
    2. 6.1 Exponential Functions
    3. 6.2 Graphs of Exponential Functions
    4. 6.3 Logarithmic Functions
    5. 6.4 Graphs of Logarithmic Functions
    6. 6.5 Logarithmic Properties
    7. 6.6 Exponential and Logarithmic Equations
    8. 6.7 Exponential and Logarithmic Models
    9. 6.8 Fitting Exponential Models to Data
    10. Key Terms
    11. Key Equations
    12. Key Concepts
    13. Review Exercises
    14. Practice Test
  8. 7 The Unit Circle: Sine and Cosine Functions
    1. Introduction to The Unit Circle: Sine and Cosine Functions
    2. 7.1 Angles
    3. 7.2 Right Triangle Trigonometry
    4. 7.3 Unit Circle
    5. 7.4 The Other Trigonometric Functions
    6. Key Terms
    7. Key Equations
    8. Key Concepts
    9. Review Exercises
    10. Practice Test
  9. 8 Periodic Functions
    1. Introduction to Periodic Functions
    2. 8.1 Graphs of the Sine and Cosine Functions
    3. 8.2 Graphs of the Other Trigonometric Functions
    4. 8.3 Inverse Trigonometric Functions
    5. Key Terms
    6. Key Equations
    7. Key Concepts
    8. Review Exercises
    9. Practice Test
  10. 9 Trigonometric Identities and Equations
    1. Introduction to Trigonometric Identities and Equations
    2. 9.1 Solving Trigonometric Equations with Identities
    3. 9.2 Sum and Difference Identities
    4. 9.3 Double-Angle, Half-Angle, and Reduction Formulas
    5. 9.4 Sum-to-Product and Product-to-Sum Formulas
    6. 9.5 Solving Trigonometric Equations
    7. Key Terms
    8. Key Equations
    9. Key Concepts
    10. Review Exercises
    11. Practice Test
  11. 10 Further Applications of Trigonometry
    1. Introduction to Further Applications of Trigonometry
    2. 10.1 Non-right Triangles: Law of Sines
    3. 10.2 Non-right Triangles: Law of Cosines
    4. 10.3 Polar Coordinates
    5. 10.4 Polar Coordinates: Graphs
    6. 10.5 Polar Form of Complex Numbers
    7. 10.6 Parametric Equations
    8. 10.7 Parametric Equations: Graphs
    9. 10.8 Vectors
    10. Key Terms
    11. Key Equations
    12. Key Concepts
    13. Review Exercises
    14. Practice Test
  12. 11 Systems of Equations and Inequalities
    1. Introduction to Systems of Equations and Inequalities
    2. 11.1 Systems of Linear Equations: Two Variables
    3. 11.2 Systems of Linear Equations: Three Variables
    4. 11.3 Systems of Nonlinear Equations and Inequalities: Two Variables
    5. 11.4 Partial Fractions
    6. 11.5 Matrices and Matrix Operations
    7. 11.6 Solving Systems with Gaussian Elimination
    8. 11.7 Solving Systems with Inverses
    9. 11.8 Solving Systems with Cramer's Rule
    10. Key Terms
    11. Key Equations
    12. Key Concepts
    13. Review Exercises
    14. Practice Test
  13. 12 Analytic Geometry
    1. Introduction to Analytic Geometry
    2. 12.1 The Ellipse
    3. 12.2 The Hyperbola
    4. 12.3 The Parabola
    5. 12.4 Rotation of Axes
    6. 12.5 Conic Sections in Polar Coordinates
    7. Key Terms
    8. Key Equations
    9. Key Concepts
    10. Review Exercises
    11. Practice Test
  14. 13 Sequences, Probability, and Counting Theory
    1. Introduction to Sequences, Probability and Counting Theory
    2. 13.1 Sequences and Their Notations
    3. 13.2 Arithmetic Sequences
    4. 13.3 Geometric Sequences
    5. 13.4 Series and Their Notations
    6. 13.5 Counting Principles
    7. 13.6 Binomial Theorem
    8. 13.7 Probability
    9. Key Terms
    10. Key Equations
    11. Key Concepts
    12. Review Exercises
    13. Practice Test
  15. A | Proofs, Identities, and Toolkit Functions
  16. Answer Key
    1. Chapter 1
    2. Chapter 2
    3. Chapter 3
    4. Chapter 4
    5. Chapter 5
    6. Chapter 6
    7. Chapter 7
    8. Chapter 8
    9. Chapter 9
    10. Chapter 10
    11. Chapter 11
    12. Chapter 12
    13. Chapter 13
  17. Index

Learning Objectives

In this section you will:
  • Solve quadratic equations by factoring.
  • Solve quadratic equations by the square root property.
  • Solve quadratic equations by completing the square.
  • Solve quadratic equations by using the quadratic formula.
Two televisions side-by-side. The right television is slightly larger than the left.
Figure 1

The computer monitor on the left in Figure 1 is a 23.6-inch model and the one on the right is a 27-inch model. Proportionally, the monitors appear very similar. If there is a limited amount of space and we desire the largest monitor possible, how do we decide which one to choose? In this section, we will learn how to solve problems such as this using four different methods.

Solving Quadratic Equations by Factoring

An equation containing a second-degree polynomial is called a quadratic equation. For example, equations such as 2 x 2 +3x1=0 2 x 2 +3x1=0and x 2 4=0 x 2 4=0are quadratic equations. They are used in countless ways in the fields of engineering, architecture, finance, biological science, and, of course, mathematics.

Often the easiest method of solving a quadratic equation is factoring. Factoring means finding expressions that can be multiplied together to give the expression on one side of the equation.

If a quadratic equation can be factored, it is written as a product of linear terms. Solving by factoring depends on the zero-product property, which states that if ab=0, ab=0, then a=0 a=0 or b=0, b=0, where a and b are real numbers or algebraic expressions. In other words, if the product of two numbers or two expressions equals zero, then one of the numbers or one of the expressions must equal zero because zero multiplied by anything equals zero.

Multiplying the factors expands the equation to a string of terms separated by plus or minus signs. So, in that sense, the operation of multiplication undoes the operation of factoring. For example, expand the factored expression ( x2 )( x+3 ) ( x2 )( x+3 )by multiplying the two factors together.

( x2 )( x+3 ) = x 2 +3x2x6 = x 2 +x6 ( x2 )( x+3 ) = x 2 +3x2x6 = x 2 +x6

The product is a quadratic expression. Set equal to zero, x 2 +x6=0 x 2 +x6=0 is a quadratic equation. If we were to factor the equation, we would get back the factors we multiplied.

The process of factoring a quadratic equation depends on the leading coefficient, whether it is 1 or another integer. We will look at both situations; but first, we want to confirm that the equation is written in standard form, a x 2 +bx+c=0, a x 2 +bx+c=0, where a, b, and c are real numbers, and a0. a0. The equation x 2 +x6=0 x 2 +x6=0is in standard form.

We can use the zero-product property to solve quadratic equations in which we first have to factor out the greatest common factor (GCF), and for equations that have special factoring formulas as well, such as the difference of squares, both of which we will see later in this section.

The Zero-Product Property and Quadratic Equations

The zero-product property states

If ab=0, then a=0 or b=0, If ab=0, then a=0 or b=0,

where a and b are real numbers or algebraic expressions.

A quadratic equation is an equation containing a second-degree polynomial; for example

a x 2 +bx+c=0 a x 2 +bx+c=0

where a, b, and c are real numbers, and if a0, a0, it is in standard form.

Solving Quadratics with a Leading Coefficient of 1

In the quadratic equation x 2 +x6=0, x 2 +x6=0, the leading coefficient, or the coefficient of x 2 , x 2 , is 1. We have one method of factoring quadratic equations in this form.

How To

Given a quadratic equation with the leading coefficient of 1, factor it.

  1. Find two numbers whose product equals c and whose sum equals b.
  2. Use those numbers to write two factors of the form ( x+k ) or ( xk ), ( x+k ) or ( xk ), where k is one of the numbers found in step 1. Use the numbers exactly as they are. In other words, if the two numbers are 1 and −2, −2, the factors are ( x+1 )( x2 ). ( x+1 )( x2 ).
  3. Solve using the zero-product property by setting each factor equal to zero and solving for the variable.

Example 1

Solving a Quadratic Equation by Factoring when the Leading Coefficient is not 1

Factor and solve the equation: x 2 +x6=0. x 2 +x6=0.

Try It #1

Factor and solve the quadratic equation: x 2 5x6=0. x 2 5x6=0.

Example 2

Solve the Quadratic Equation by Factoring

Solve the quadratic equation by factoring: x 2 +8x+15=0. x 2 +8x+15=0.

Try It #2

Solve the quadratic equation by factoring: x 2 4x21=0. x 2 4x21=0.

Example 3

Using the Zero-Product Property to Solve a Quadratic Equation Written as the Difference of Squares

Solve the difference of squares equation using the zero-product property: x 2 9=0. x 2 9=0.

Try It #3

Solve by factoring: x 2 25=0. x 2 25=0.

Solving a Quadratic Equation by Factoring when the Leading Coefficient is not 1

When the leading coefficient is not 1, we factor a quadratic equation using the method called grouping, which requires four terms. With the equation in standard form, let’s review the grouping procedures:

  1. With the quadratic in standard form, a x 2 +bx+c=0, a x 2 +bx+c=0, multiply ac. ac.
  2. Find two numbers whose product equals ac ac and whose sum equals b. b.
  3. Rewrite the equation replacing the bx bx term with two terms using the numbers found in step 1 as coefficients of x.
  4. Factor the first two terms and then factor the last two terms. The expressions in parentheses must be exactly the same to use grouping.
  5. Factor out the expression in parentheses.
  6. Set the expressions equal to zero and solve for the variable.

Example 4

Solving a Quadratic Equation Using Grouping

Use grouping to factor and solve the quadratic equation: 4 x 2 +15x+9=0. 4 x 2 +15x+9=0.

Try It #4

Solve using factoring by grouping: 12 x 2 +11x+2=0. 12 x 2 +11x+2=0.

Example 5

Solving a Polynomial of Higher Degree by Factoring

Solve the equation by factoring: −3 x 3 5 x 2 2x=0. −3 x 3 5 x 2 2x=0.

Try It #5

Solve by factoring: x 3 +11 x 2 +10x=0. x 3 +11 x 2 +10x=0.

Using the Square Root Property

When there is no linear term in the equation, another method of solving a quadratic equation is by using the square root property, in which we isolate the x 2 x 2 term and take the square root of the number on the other side of the equals sign. Keep in mind that sometimes we may have to manipulate the equation to isolate the x 2 x 2 term so that the square root property can be used.

The Square Root Property

With the x 2 x 2 term isolated, the square root property states that:

if x 2 =k,thenx=± k if x 2 =k,thenx=± k

where k is a nonzero real number.

How To

Given a quadratic equation with an x 2 x 2 term but no x x term, use the square root property to solve it.

  1. Isolate the x 2 x 2 term on one side of the equal sign.
  2. Take the square root of both sides of the equation, putting a ± ± sign before the expression on the side opposite the squared term.
  3. Simplify the numbers on the side with the ± ± sign.

Example 6

Solving a Simple Quadratic Equation Using the Square Root Property

Solve the quadratic using the square root property: x 2 =8. x 2 =8.

Example 7

Solving a Quadratic Equation Using the Square Root Property

Solve the quadratic equation: 4 x 2 +1=7. 4 x 2 +1=7.

Try It #6

Solve the quadratic equation using the square root property: 3 ( x4 ) 2 =15. 3 ( x4 ) 2 =15.

Completing the Square

Not all quadratic equations can be factored or can be solved in their original form using the square root property. In these cases, we may use a method for solving a quadratic equation known as completing the square. Using this method, we add or subtract terms to both sides of the equation until we have a perfect square trinomial on one side of the equal sign. We then apply the square root property. To complete the square, the leading coefficient, a, must equal 1. If it does not, then divide the entire equation by a. Then, we can use the following procedures to solve a quadratic equation by completing the square.

We will use the example x 2 +4x+1=0 x 2 +4x+1=0 to illustrate each step.

  1. Given a quadratic equation that cannot be factored, and with a=1, a=1, first add or subtract the constant term to the right sign of the equal sign.

    x 2 +4x=−1 x 2 +4x=−1
  2. Multiply the b term by 1 2 1 2 and square it.

    1 2 (4) = 2 2 2 = 4 1 2 (4) = 2 2 2 = 4
  3. Add ( 1 2 b ) 2 ( 1 2 b ) 2 to both sides of the equal sign and simplify the right side. We have

    x 2 +4x+4 = 1+4 x 2 +4x+4 = 3 x 2 +4x+4 = 1+4 x 2 +4x+4 = 3
  4. The left side of the equation can now be factored as a perfect square.

    x 2 +4x+4 = 3 (x+2) 2 = 3 x 2 +4x+4 = 3 (x+2) 2 = 3
  5. Use the square root property and solve.

    (x+2) 2 = ± 3 x+2 = ± 3 x = −2± 3 (x+2) 2 = ± 3 x+2 = ± 3 x = −2± 3
  6. The solutions are −2+ 3 , −2+ 3 , and −2 3 . and −2 3 .

Example 8

Solving a Quadratic by Completing the Square

Solve the quadratic equation by completing the square: x 2 3x5=0. x 2 3x5=0.

Try It #7

Solve by completing the square: x 2 6x=13. x 2 6x=13.

Using the Quadratic Formula

The fourth method of solving a quadratic equation is by using the quadratic formula, a formula that will solve all quadratic equations. Although the quadratic formula works on any quadratic equation in standard form, it is easy to make errors in substituting the values into the formula. Pay close attention when substituting, and use parentheses when inserting a negative number.

We can derive the quadratic formula by completing the square. We will assume that the leading coefficient is positive; if it is negative, we can multiply the equation by −1 −1 and obtain a positive a. Given a x 2 +bx+c=0, a x 2 +bx+c=0, a0, a0, we will complete the square as follows:

  1. First, move the constant term to the right side of the equal sign:

    a x 2 +bx=c a x 2 +bx=c
  2. As we want the leading coefficient to equal 1, divide through by a:

    x 2 + b a x= c a x 2 + b a x= c a
  3. Then, find 1 2 1 2 of the middle term, and add ( 1 2 b a ) 2 = b 2 4 a 2 ( 1 2 b a ) 2 = b 2 4 a 2 to both sides of the equal sign:

    x 2 + b a x+ b 2 4 a 2 = b 2 4 a 2 c a x 2 + b a x+ b 2 4 a 2 = b 2 4 a 2 c a
  4. Next, write the left side as a perfect square. Find the common denominator of the right side and write it as a single fraction:

    ( x+ b 2a ) 2 = b 2 4ac 4 a 2 ( x+ b 2a ) 2 = b 2 4ac 4 a 2
  5. Now, use the square root property, which gives

    x+ b 2a = ± b 2 4ac 4 a 2 x+ b 2a = ± b 2 4ac 2a x+ b 2a = ± b 2 4ac 4 a 2 x+ b 2a = ± b 2 4ac 2a
  6. Finally, add b 2a b 2a to both sides of the equation and combine the terms on the right side. Thus,

    x= b± b 2 4ac 2a x= b± b 2 4ac 2a

The Quadratic Formula

Written in standard form, a x 2 +bx+c=0, a x 2 +bx+c=0, any quadratic equation can be solved using the quadratic formula:

x= b± b 2 4ac 2a x= b± b 2 4ac 2a

where a, b, and c are real numbers and a0. a0.

How To

Given a quadratic equation, solve it using the quadratic formula

  1. Make sure the equation is in standard form: a x 2 +bx+c=0. a x 2 +bx+c=0.
  2. Make note of the values of the coefficients and constant term, a,b, a,b,and c. c.
  3. Carefully substitute the values noted in step 2 into the equation. To avoid needless errors, use parentheses around each number input into the formula.
  4. Calculate and solve.

Example 9

Solve the Quadratic Equation Using the Quadratic Formula

Solve the quadratic equation: x 2 +5x+1=0. x 2 +5x+1=0.

Example 10

Solving a Quadratic Equation with the Quadratic Formula

Use the quadratic formula to solve x 2 +x+2=0. x 2 +x+2=0.

Try It #8

Solve the quadratic equation using the quadratic formula: 9 x 2 +3x2=0. 9 x 2 +3x2=0.

The Discriminant

The quadratic formula not only generates the solutions to a quadratic equation, it tells us about the nature of the solutions when we consider the discriminant, or the expression under the radical, b 2 4ac. b 2 4ac.The discriminant tells us whether the solutions are real numbers or complex numbers, and how many solutions of each type to expect. Table 1 relates the value of the discriminant to the solutions of a quadratic equation.

Value of Discriminant Results
b 2 4ac=0 b 2 4ac=0 One rational solution (double solution)
b 2 4ac>0, b 2 4ac>0,perfect square Two rational solutions
b 2 4ac>0, b 2 4ac>0,not a perfect square Two irrational solutions
b 2 4ac<0 b 2 4ac<0 Two complex solutions
Table 1

The Discriminant

For a x 2 +bx+c=0 a x 2 +bx+c=0 , where a a , b b , and c c are real numbers, the discriminant is the expression under the radical in the quadratic formula: b 2 4ac. b 2 4ac.It tells us whether the solutions are real numbers or complex numbers and how many solutions of each type to expect.

Example 11

Using the Discriminant to Find the Nature of the Solutions to a Quadratic Equation

Use the discriminant to find the nature of the solutions to the following quadratic equations:

  1. x 2 +4x+4=0 x 2 +4x+4=0
  2. 8 x 2 +14x+3=0 8 x 2 +14x+3=0
  3. 3 x 2 5x2=0 3 x 2 5x2=0
  4. 3 x 2 10x+15=0 3 x 2 10x+15=0

Using the Pythagorean Theorem

One of the most famous formulas in mathematics is the Pythagorean Theorem. It is based on a right triangle, and states the relationship among the lengths of the sides as a 2 + b 2 = c 2 , a 2 + b 2 = c 2 , where a a and b b refer to the legs of a right triangle adjacent to the 90° 90°angle, and c c refers to the hypotenuse. It has immeasurable uses in architecture, engineering, the sciences, geometry, trigonometry, and algebra, and in everyday applications.

We use the Pythagorean Theorem to solve for the length of one side of a triangle when we have the lengths of the other two. Because each of the terms is squared in the theorem, when we are solving for a side of a triangle, we have a quadratic equation. We can use the methods for solving quadratic equations that we learned in this section to solve for the missing side.

The Pythagorean Theorem is given as

a 2 + b 2 = c 2 a 2 + b 2 = c 2

where a a and b b refer to the legs of a right triangle adjacent to the 90 90 angle, and c c refers to the hypotenuse, as shown in Figure 4.

Right triangle with the base labeled: a, the height labeled: b, and the hypotenuse labeled: c
Figure 4

Example 12

Finding the Length of the Missing Side of a Right Triangle

Find the length of the missing side of the right triangle in Figure 5.

Right triangle with the base labeled: a, the height labeled: 4, and the hypotenuse labeled 12.
Figure 5
Try It #9

Use the Pythagorean Theorem to solve the right triangle problem: Leg a measures 4 units, leg b measures 3 units. Find the length of the hypotenuse.

2.5 Section Exercises

Verbal

1.

How do we recognize when an equation is quadratic?

2.

When we solve a quadratic equation, how many solutions should we always start out seeking? Explain why when solving a quadratic equation in the form a x 2 +bx+c=0 a x 2 +bx+c=0 we may graph the equation y=a x 2 +bx+c y=a x 2 +bx+c and have no zeroes (x-intercepts).

3.

When we solve a quadratic equation by factoring, why do we move all terms to one side, having zero on the other side?

4.

In the quadratic formula, what is the name of the expression under the radical sign b 2 4ac, b 2 4ac, and how does it determine the number of and nature of our solutions?

5.

Describe two scenarios where using the square root property to solve a quadratic equation would be the most efficient method.

Algebraic

For the following exercises, solve the quadratic equation by factoring.

6.

x 2 +4x21=0 x 2 +4x21=0

7.

x 2 9x+18=0 x 2 9x+18=0

8.

2 x 2 +9x5=0 2 x 2 +9x5=0

9.

6 x 2 +17x+5=0 6 x 2 +17x+5=0

10.

4 x 2 12x+8=0 4 x 2 12x+8=0

11.

3 x 2 75=0 3 x 2 75=0

12.

8 x 2 +6x9=0 8 x 2 +6x9=0

13.

4 x 2 =9 4 x 2 =9

14.

2 x 2 +14x=36 2 x 2 +14x=36

15.

5 x 2 =5x+30 5 x 2 =5x+30

16.

4 x 2 =5x 4 x 2 =5x

17.

7 x 2 +3x=0 7 x 2 +3x=0

18.

x 3 9 x =2 x 3 9 x =2

For the following exercises, solve the quadratic equation by using the square root property.

19.

x 2 =36 x 2 =36

20.

x 2 =49 x 2 =49

21.

( x1 ) 2 =25 ( x1 ) 2 =25

22.

( x3 ) 2 =7 ( x3 ) 2 =7

23.

( 2x+1 ) 2 =9 ( 2x+1 ) 2 =9

24.

( x5 ) 2 =4 ( x5 ) 2 =4

For the following exercises, solve the quadratic equation by completing the square. Show each step.

25.

x 2 9x22=0 x 2 9x22=0

26.

2 x 2 8x5=0 2 x 2 8x5=0

27.

x 2 6x=13 x 2 6x=13

28.

x 2 + 2 3 x 1 3 =0 x 2 + 2 3 x 1 3 =0

29.

2+z=6 z 2 2+z=6 z 2

30.

6 p 2 +7p20=0 6 p 2 +7p20=0

31.

2 x 2 3x1=0 2 x 2 3x1=0

For the following exercises, determine the discriminant, and then state how many solutions there are and the nature of the solutions. Do not solve.

32.

2 x 2 6x+7=0 2 x 2 6x+7=0

33.

x 2 +4x+7=0 x 2 +4x+7=0

34.

3 x 2 +5x8=0 3 x 2 +5x8=0

35.

9 x 2 30x+25=0 9 x 2 30x+25=0

36.

2 x 2 3x7=0 2 x 2 3x7=0

37.

6 x 2 x2=0 6 x 2 x2=0

For the following exercises, solve the quadratic equation by using the quadratic formula. If the solutions are not real, state No Real Solution.

38.

2 x 2 +5x+3=0 2 x 2 +5x+3=0

39.

x 2 +x=4 x 2 +x=4

40.

2 x 2 8x5=0 2 x 2 8x5=0

41.

3 x 2 5x+1=0 3 x 2 5x+1=0

42.

x 2 +4x+2=0 x 2 +4x+2=0

43.

4+ 1 x 1 x 2 =0 4+ 1 x 1 x 2 =0

Technology

For the following exercises, enter the expressions into your graphing utility and find the zeroes to the equation (the x-intercepts) by using 2nd CALC 2:zero. Recall finding zeroes will ask left bound (move your cursor to the left of the zero,enter), then right bound (move your cursor to the right of the zero,enter), then guess (move your cursor between the bounds near the zero, enter). Round your answers to the nearest thousandth.

44.

Y 1 =4 x 2 +3x2 Y 1 =4 x 2 +3x2

45.

Y 1 =−3 x 2 +8x1 Y 1 =−3 x 2 +8x1

46.

Y 1 =0.5 x 2 +x7 Y 1 =0.5 x 2 +x7

47.

To solve the quadratic equation x 2 +5x7=4, x 2 +5x7=4,we can graph these two equations

Y 1 = x 2 +5x7 Y 2 =4 Y 1 = x 2 +5x7 Y 2 =4

and find the points of intersection. Recall 2nd CALC 5:intersection. Do this and find the solutions to the nearest tenth.

48.

To solve the quadratic equation 0.3 x 2 +2x4=2, 0.3 x 2 +2x4=2,we can graph these two equations

Y 1 =0.3 x 2 +2x4 Y 2 =2 Y 1 =0.3 x 2 +2x4 Y 2 =2

and find the points of intersection. Recall 2nd CALC 5:intersection. Do this and find the solutions to the nearest tenth.

Extensions

49.

Beginning with the general form of a quadratic equation, a x 2 +bx+c=0, a x 2 +bx+c=0, solve for x by using the completing the square method, thus deriving the quadratic formula.

50.

Show that the sum of the two solutions to the quadratic equation is b a b a .

51.

A person has a garden that has a length 10 feet longer than the width. Set up a quadratic equation to find the dimensions of the garden if its area is 119 ft.2. Solve the quadratic equation to find the length and width.

52.

Abercrombie and Fitch stock had a price given as P=0.2 t 2 5.6t+50.2, P=0.2 t 2 5.6t+50.2,where t t is the time in months from 1999 to 2001. ( t=1 t=1 is January 1999). Find the two months in which the price of the stock was $30.

53.

Suppose that an equation is given p=−2 x 2 +280x1000, p=−2 x 2 +280x1000,where x x represents the number of items sold at an auction and p p is the profit made by the business that ran the auction. How many items sold would make this profit a maximum? Solve this by graphing the expression in your graphing utility and finding the maximum using 2nd CALC maximum. To obtain a good window for the curve, set x x [0,200] and y y [0,10000].

Real-World Applications

54.

A formula for the normal systolic blood pressure for a man age A, A, measured in mmHg, is given as P=0.006 A 2 0.02A+120. P=0.006 A 2 0.02A+120.Find the age to the nearest year of a man whose normal blood pressure measures 125 mmHg.

55.

The cost function for a certain company is C=60x+300 C=60x+300 and the revenue is given by R=100x0.5 x 2 . R=100x0.5 x 2 .Recall that profit is revenue minus cost. Set up a quadratic equation and find two values of x (production level) that will create a profit of $300.

56.

A falling object travels a distance given by the formula d=5t+16 t 2 d=5t+16 t 2 ft, where t t is measured in seconds. How long will it take for the object to traveled 74 ft?

57.

A vacant lot is being converted into a community garden. The garden and the walkway around its perimeter have an area of 378 ft2. Find the width of the walkway if the garden is 12 ft. wide by 15 ft. long.

A rectangle inside of a larger rectangle. The smaller rectangle has the length labeled: 15 feet and the width labeled: 12 feet. The distance between the two rectangles is labeled as x on all four sides.
58.

An epidemiological study of the spread of a certain influenza strain that hit a small school population found that the total number of students, P, P, who contracted the flu t t days after it broke out is given by the model P= t 2 +13t+130, P= t 2 +13t+130, where 1t6. 1t6. Find the day that 160 students had the flu. Recall that the restriction on t t is at most 6.

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