Copyright © Cengage Learning. All rights reserved. 4 Quadratic Functions

Copyright © Cengage Learning. All rights reserved. 4.6 Solving Quadratic Equations by Using the Quadratic Formula

3 Objectives  Solve a quadratic equation using the quadratic formula.  Solve systems of equations with quadratics.

4 Solving by the Quadratic Formula

5 If we start with a standard quadratic form, we can solve for x while leaving the coefficients and constants unknown. The only requirement is that a  0, or the division would be undefined and the equation would not be a quadratic. ax 2 + bx + c = 0 ax 2 + bx = –c We start with the standard quadratic 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 cannot equal zero, so that division by a is defined.

6 Solving by the Quadratic Formula Factor and simplify. Use the square root property. Add a constant that will complete the square (half of the coefficient of x, squared).

7 Solving by the Quadratic Formula Get x by itself. Simplify the denominator of the radical.

8 Solving by the Quadratic Formula Because we solved the quadratic equation for any values for a, b, and c, the mathematics will be true for any real values of a  0, b, and c. Therefore, in general, we have solved any quadratic from standard form. The final result is called the quadratic formula. This formula can be used to solve any quadratic equation in standard form as long as the equation equals zero.

9 Solving by the Quadratic Formula

10 Example 1 – Using the quadratic formula Solve the following quadratic equations. Round your answers to three decimal places. Check your answers. a. 3x 2 – 6x – 24 = 0 b. 4t 2 – 8t + 5 = 50 c. 3.4x x – 7.8 = 0 Solution: a. 3x 2 – 6x – 24 = 0 The equation is in standard form and equal to zero, so use the quadratic formula.

11 Example 1 – Solution Separate into two equations. Simplify. cont’d

12 Example 1 – Solution x = 4 x = –2 3(4) 2 – 6(4) – 24 ≟ 0 48 – 24 – 24 ≟ 0 0 = 0 3(–2) 2 – 6(–2) – 24 ≟ – 24 ≟ 0 0 = 0 Both answers work. Check both answers. cont’d

13 Example 1 – Solution b. 4t 2 – 8t + 5 = 50 4t 2 – 8t – 45 = 0 Use the quadratic formula. Set the equation equal to zero. cont’d

14 Example 1 – Solution Simplify Separate into two equations cont’d Check both answers using the table. Both answers work.

15 Example 1 – Solution c. 3.4x x – 7.8 = 0 Separate into two equations. Simplify. cont’d

16 Example 1 – Solution x  x  –2.253 cont’d Check both answers using the graph. Both answers work.

17 Determining Which Algebraic Method to Use When Solving a Quadratic Equation

18 Determining Which Algebraic Method to Use When Solving a Quadratic Equation When we are faced with an equation to solve, we should first decide which type of equation we are trying to solve (linear, quadratic, or higher-degree polynomial) and then choose a method that is best suited to solve that equation. We have looked at four algebraic methods to solve a quadratic equation: Square root property Completing the square

19 Determining Which Algebraic Method to Use When Solving a Quadratic Equation Factoring Quadratic formula Determining what method is best depends on the characteristics of the equation we are trying to solve.

20 Determining Which Algebraic Method to Use When Solving a Quadratic Equation Square Root Property The square root property works best when the quadratic is in vertex form or when there is a squared variable term but no other variable terms. (x + 5) 2 – 9 = 16 x = 49 6x 2 – 4 = 18 In all of these equations, the square root property would be a good method to use.

21 Determining Which Algebraic Method to Use When Solving a Quadratic Equation Remember, when solving using the square root property, to isolate the squared variable expression on one side before using the square root property. Also do not forget to use the plus/minus symbol to indicate all possible answers. Completing the square Completing the square works well for equations that have both a squared term and a first-degree term. This method is usually easiest if the numbers are not too large and the leading coefficient is 1.

22 Determining Which Algebraic Method to Use When Solving a Quadratic Equation x 2 + 4x + 9 = 0 x 2 – 5x = 20 After completing the square, we will again use the square root property to solve. Factoring Factoring works best when the numbers are not too large or when the terms are higher than second degree. Always remember to first factor out the greatest common factor. Although factoring does not work with all quadratics, if the equation factors easily, it can be one of the fastest solution methods.

23 Determining Which Algebraic Method to Use When Solving a Quadratic Equation For factoring to be used when solving, the equation must be equal to zero so that the zero product property can be used. x 2 + 5x + 6 = 0 x 3 + x 2 – 6x = 0 Equations that cannot be factored may still have solutions, so use one of the other methods, such as completing the square or the quadratic formula. Occasionally, an equation can be factored partially, and then the separate pieces can be solved using another method.

24 Determining Which Algebraic Method to Use When Solving a Quadratic Equation Quadratic Formula The quadratic formula will work with any quadratic, but it is easiest if the quadratic starts out in standard form. Because the quadratic formula basically requires substituting values for a, b, and c and then simplifying an arithmetic expression, the formula will work equally well for large or small numbers. Whenever a quadratic equation has decimals or fractions, the quadratic formula is probably the best method to choose.

25 Determining Which Algebraic Method to Use When Solving a Quadratic Equation 5x x – 85 = x 2 – 3.4x + 9 = 0 In solving with the square root property or the quadratic formula, a negative under the square root will indicate no real solutions.

26 Example 3 – Choosing a method to solve equations Solve the following equations using any method you wish. Check your answers using the calculator table or graph. a. 5x x 2 – 5x = –19x b. x x + 7 = 0 c. 5x 2 – 30 = 40 d. 2x(x + 5) = 4x e. x 3 + 8x 2 + 5x = 0

27 Example 3(a) – Solution This equation has a third-degree term, so we will try to factor it and break it up into smaller pieces that will be easier to solve. The first step will be to set one side of the equation equal to zero. 5x x 2 – 5x = –19x 5x x x = 0 x(5x x + 14) = 0 x(5x 2 + 2x + 35x + 14) = 0 cont’d Set the equation equal to zero. Factor.

28 Example 3(a) – Solution x[(5x 2 + 2x) + (35x + 14)] = 0 x[x(5x + 2) + 7(5x + 2)] = 0 x(x + 7)(5x + 2) = 0 x = 0 x + 7 = 0 5x + 2 = 0 x = 0 x = –7 x = cont’d Check all the three answers using the table All the answers work. Separate into three equations and solve.

29 Example 3(b) – Solution This equation does not factor (try it!), so we must use another method. Since the equation has a squared term and a first-degree term, we will not be able to use the square root property directly. Therefore, we could complete the square and then use the square root property. x x + 7 = 0 x x = –7 x x + 36 = – (x + 6) 2 = 29 cont’d Complete the square.

30 Example 3(b) – Solution cont’d Check all the three answers using the graph Both answers work. Use the square root property. Separate into two equations and solve.

31 Example 3(c) – Solution This equation has no first-degree term, so the square root property will be a good method to choose. First isolate the squared variable expression. 5x 2 – 30 = 40 5x 2 = 70 x 2 = 14 cont’d Use the square root property. Separate into two equations and solve. Isolate the squared variable expression.

32 Example 3(c) – Solution cont’d Both answers work. Check both the answers using the table

33 Example 3(d) – Solution This equation looks like it is already factored. Since it does not equal zero, we will have to multiply it out and set it equal to zero before we can factor and solve. 2x(x + 5) = 4x 2x x = 4x 2x 2 + 6x = 0 2x(x + 3) = 0 2x = 0 x + 3 = 0 x = 0 x = –3 cont’d Multiply out using the distributive property. Separate into two equations and solve. Set the equation equal to zero. Factor.

34 Example 3(d) – Solution cont’d Both answers work. Check both the answers using the table

35 Example 3(e) – Solution This equation has a third-degree term, so it should be factored. x 3 + 8x 2 + 5x = 0 x(x 2 + 8x + 5) = 0 The equation cannot be factored any further. We will set each factor equal to zero and then solve the remaining quadratic equation using the quadratic formula. cont’d Factor out the common term.

36 Example 3(e) – Solution x = 0 or x 2 + 8x + 5 = 0 a = 1 b = 8 c = 5 cont’d Separate into two equations and solve. Use the Quadratic Formula. Simplify the radical.

37 Example 3(e) – Solution This equation appears to have three answers. Check them in the table. cont’d All the answers work.

38 Solving Systems of Equations with Quadratics

39 Solving Systems of Equations with Quadratics Let us consider systems of equations that contain quadratics. We have learned that we could solve systems using three different methods: graphing, substitution, and elimination. Using graphs and tables to estimate solutions to systems of equations that involve functions other than lines is the same process as with linear systems. When using the graph, we look for the place(s) where the two graphs intersect.

40 Solving Systems of Equations with Quadratics When using a table, we find input value(s) that make the outputs for both equations equal. When looking for these input value(s), notice when one equation changes from being smaller than the other equation to being larger than the other. The intersection of the two equations must be between these input values.

41 Example 4 – Using graphs and tables to solve systems of equations Use the given graph or table to estimate the solutions to the systems of equations.

42 Example 4 – Solution a. From the graph, we can see that the two graphs intersect at about the points (4, 2) and (–2, 14). b. The table shows that the two equations are equal at the points (2, 10) and (5, 64). c. This table does not show any exact places where the equations are equal. The equation in Y1 is greater than Y2 at x = –4.4, but Y1 is less than Y2 at x = – So there must be an intersection between these values. One estimate for this intersection could be (–4.38, 0.3).

43 Example 4 – Solution The two equations must also intersect between x = –0.12 and x = –0.11, so we might estimate this intersection at about (–0.115, 2.48). Remember that these are only estimates. Many other answers could be reasonable. cont’d

44 Solving Systems of Equations with Quadratics Although graphs and tables can be used to find solutions to these systems, we can also use algebraic methods to solve. In solving systems that contain quadratics algebraically, substitution is often the best choice.

45 Example 5 – Solving systems of equations algebraically Solve the following systems of equations. a. y = 4x + 9 b. y = 5x 2 + 2x + 7 y = x 2 + 5x + 3 y = 2x 2 – 3x + 10 Solution: a. y = 4x + 9 y = x 2 + 5x + 3 Substitute 4x + 9 for y in the second equation and solve for x. 4x + 9 = x 2 + 5x + 3 Substitute for y.

46 Example 5 – Solution 0 = x 2 + x – 6 0 = (x + 3)(x – 2) x + 3 = 0 x – 2 = 0 x = –3 x = 2 y = 4(–3) + 9 y = –3 y = 4(2) + 9 y = 17 (–3, –3) (2, 17) Separate into two equations and solve. Use the values of x to find the corresponding values of y. These two points are the solutions to the system of equations. cont’d Set the equation equal to zero. Factor.

47 Example 5 – Solution These answers can be checked by using either the table feature or the graph feature on the calculator. First, put both equations into the Y = screen, and then graph them using a window that will include both answers. Another option is to go to the table and input both x-values to see whether the y-values are the same. Remember that some rounding error can occur if the solutions were not exact. cont’d

48 Example 5 – Solution b.y = 5x 2 + 2x + 7 y = 2x 2 – 3x + 10 Substitute for y and solve for x. 5x 2 + 2x + 7 = 2x 2 – 3x x 2 + 5x –3 = 0 a = 3 b = 5 c = –3 Substitute for y and simplify. You are left with a quadratic equation that can be solved by using the quadratic formula. cont’d

49 Example 5 – Solution Use the x-values to find the corresponding y-values. y = 5(0.468) 2 + 2(0.468) + 7 cont’d

50 Example 5 – Solution y = y = 5(–2.135) 2 + 2(–2.135) + 7 y = (0.468, 9.031) (–2.135, ) Check the answers using the table or graph. cont’d There are two answers to this system Both answers work.

51 Solving Systems of Equations with Quadratics In using the quadratic formula, the value of the discriminant, b 2 – 4ac, determines whether the equation will have real solutions or not. When the discriminant is negative, the quadratic equation will have no real solutions. In the real number system, we cannot take the square root of negative numbers.