B.3 Solving Equations Algebraically and Graphically

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1 B.3 Solving Equations Algebraically and Graphically 1

2 Equations and Solutions of Equations An equation in x is a statement that two algebraic expressions are equal. To solve an equation in x means to find all values of x for which the equation is true. Such values are solutions. For example, x = 4 is a solution of the equation 3x - 5 = 7 because 3(4) - 5 = 7 is a true statement. An equation that is true for just some (or even none) of the real numbers in the domain of the variable is called a conditional equation. The equation x2-9 = 0 is conditional because x = 3 and x = -3 are the only values in the domain that satisfy the equation. 2

3 Solving an Equation Involving Fractions Example 1 Solve: x + 3x =

4 Solving an Equation Involving Fractions Example 1 Solve x + 3x = Multiply each term by the LCD of 12 (12) x + (12) 3x = (12) Divide out and multiply Combine like terms Divide each side by 13 4x + 9x = 24 13x = 24 x = Study Tip: after solving an equation, check each solution in the original equation.

5 Extraneous Solutions When multiplying or dividing an equation by a variable expression, it is possible to introduce an extraneous solution - one that does not satisfy the original equation. What do I mean by a variable expression? 5

6 An Equation with an Extraneous Solution Example 2 Solve: 1 x-2 = 3 x+2-6x x2-4 6

7 An Equation with an Extraneous Solution Example 2 Algebraic Solution 1 = x-2 3 x+2-6x x2-4 The LCD is x2-4 = (x + 2)(x - 2) (x + 2)(x - 2) 1 = x-2 (x + 2) = 3(x - 2) - 6x x + 2 = 3x - 6-6x 4x = -8 x = -2 (x + 2)(x - 2) 3 x+2 - (x + 2)(x - 2) 6x x2-4 A check of x = -2 in the original equation shows that it yields a denominator of zero. So, x = -2 is an extraneous solution, and the original equation has no solution. 7

8 An Equation with an Extraneous Solution Example 2 Graphical Solution: Use a graphing calculator to graph the left and right sides of the equation. y1 = 1 x-2 y2 = x+2 3-6x x2-4 The graphs of the equations do not appear to intersect. This means that there is no point for which the left side of the equation y1 is equal to the right side of the equation y2. So, the equation appears to have no solutions. 8

9 Intercepts and Solutions Definition of Intercepts 1. The point (a, 0) is called an x-intercept of the graph of an equation if it is a solution point of the equation. To find the x-intercept(s), set y equal to 0 and solve the equation for x. 2. The point (0, b) is called a y-intercept of the graph of an equation if it is a solution point of the equation. To find the y-intercept(s), set x equal to 0 and solve the equation for y. 9

10 Finding x- and y-intercepts Example 3 Find the x- and y-intercepts of the graph of 2x + 3y = 5 10

11 Finding x- and y-intercepts Example 3 Solution 2x + 3y = 5 To find the x-intercept, let y = 0 and solve for x 2x + 3(0) = 5 2x = 5 x = 5/2 which implies that the graph has one x-intercept: (5/2, 0) To find the y-intercept, let x = 0 and solve for y 2(0) + 3y = 5 3y = 5 y = 5/3 which implies that the graph has one y-intercept: (0, 5/3) 11

12 Finding Solutions Graphically Graphical Approximations of Solutions of an Equation 1. Write the equation in general form, y = 0, with the nonzero terms on one side of the equation and zero on the other side. 2. Use a graphing calculator to graph the equation. Be sure the viewing window shows all the relevant features of the graph. 3. Use the zero or root feature or the zoom and trace features of the graphing calculator to approximate the x-intercepts of the graph. 12

13 Finding Solutions of an Equation Graphically Example 4 Use a graphing calculator to approximate the solutions of 2x3-3x + 2 = 0 13

14 Finding Solutions of an Equation Graphically Solution: 2x3-3x + 2 = 0 Graph the function y = 2x3-3x + 2 You can see that there is one x-intercept between -2 and -1, at approximately By using the zero or root feature, you can improve the approximation. Choose a left bound of x = -2, and right bound of x = -1. The solution is x Check this approximation in your calculator. 14

15 Points of Intersection of Two Graphs An ordered pair that is a solution of two different equations if called a point of intersection of the graphs of two equations. To find the points of intersection of the graphs of two equations, solve each equation for y (or x) and set the two results equal to each other. The resulting equation will be an equation in one variable that can be solved using standard procedures. 15

16 Finding Points of Intersection Example 6 Find the points of intersection of the graphs of 2x - 3y = -2 and 4x - y = 6 16

17 Finding Points of Intersection Algebraic Solution: Solve each equation for y 2x - 3y = -2 4x - y = 6-3y = -2x - 2 4x = 6 + y y = (⅔)x + ⅔ 4x - 6 = y Next, set the two expressions for y equal to each other and solve the resulting equation for x, (⅔)x + ⅔ = 4x - 6 (3)[(⅔)x + ⅔] = (3)[4x - 6] 2x + 2 = 12x - 18 When x = 2, the y-value of each of the original equations is 2. So, the point of intersection is (2, 2). 20 = 10x 2=x 17

18 Finding Points of Intersection Graphical Solution Solve each equation for y y = (⅔)x + ⅔ and y = 4x - 6 Use a graphing calculator to graph both equations. Use the intersect feature to approximate the point of intersections to be (2, 2). 18

19 Solving an Equation of Quadratic Type Example 8 Solve x4-3x2 + 2 = 0 We say an expression is said to be in quadratic form when it is written in the form of ax2 + bx + c Can we write this in this form? Then factor? 19

20 Solving an Equation of Quadratic Type Example 8 Solve x4-3x2 + 2 = 0 This can be written in quadratic form of au2 + bu + c, when u = x2. We can use factoring to solve the equation. (u)2-3(u) + 2 = 0 (u - 1)(u - 2) = 0 (x2-1)(x2-2) = 0 Let u = x2 Factor Plug x2 back in (x + 1)(x - 1)(x2-2) = 0 x+1=0 x = -1 x-1=0 x=1 x2-2 = 0 Check all four solutions algebraically x = +/- 2 20

21 Solving an Equation of Quadratic Type Example 8 Solve x4-3x2 + 2 = 0 After checking each solution, you see that they are all valid. 21

22 Solving an Equation Involving a Radical Example 10 Solve (2x + 7) - x = 2 22

23 Solving an Equation Involving a Radical Solution: Solve (2x + 7) - x = 2 (2x + 7) = x + 2 Isolate the radical [ (2x + 7)]2 = (x + 2)2 Square each side 2x + 7 = x2 + 4x + 4 Write in general form 0 = x2 + 2x - 3 Factor 0 = (x + 3)(x - 1) 0=x+3 x = -3 0=x-1 x=1 Substitute each solution back into the original equation to check to see whether there is an extraneous solution. 23

24 Solving an Equation Involving a Radical Solution: (2x + 7) - x = 2 (2(-3) + 7) - (-3) =? 2 (1) + 3 =? =? So x = -3 is an extraneous solution (2x + 7) - x = 2 (2(1) + 7) - (1) = 2 (9) - 1 = 2 3-1=2 So x = 1 is valid, and the only real solution. 24

25 Solving an Equation Involving a Radical Graphical Solution: (2x + 7) - x = 2 First rewrite the equation as (2x + 7) - x - 2 = 0 Now graph y = (2x + 7) - x - 2 Notice that the domain is x - 7/2 because the expression under the radical cannot be negative. 25

26 Solving an Equation Involving Two Radicals Example 11 (2x + 6) - (x + 4) = 1 26

27 Solving an Equation Involving Two Radicals Solution: (2x + 6) - (x + 4) = 1 (2x + 6) = 1 + (x + 4) Isolate (2x + 6) [ (2x + 6)]2 = [1 + (x + 4)]2 Square each side 2x + 6 = (x + 4) + (x + 4) x + 1 = 2 (x + 4) Isolate 2 (x + 4) (x + 1)2 = [2 (x + 4)]2 Square each side x2 + 2x + 1 = 4(x + 4) x2 + 2x + 1 = 4x + 16 Write in general form x2-2x - 15 = 0 Factor (x - 5)(x + 3) = 0, so x = 5 and x = -3 Check to see if either are extraneous solutions. 27

28 Solving an Equation Involving Two Radicals Solution: (2x + 6) - (x + 4) = 1 After checking each solution, you see that the only valid solution is x = 5 28

29 Solving an Equation with Rational Exponents Example 12 Solve (x + 1)⅔ = 4 Hint: Rewrite as radical. Don t forget to check for extraneous solutions! 29

30 Solving an Equation with Rational Exponents Example 12 Solve (x + 1)⅔ = 4 (x + 1)2 = 4 ( (x + 1)2)3 = 43 Rewrite in Radical form Cube each side (x + 1)2 = 64 ((x + 1)2) = (64) Square root each side x + 1 = +/- 8 x = +/- 8-1 x = 7, x = -9 Substitute x = 7 and x = -9 into the original equation to determine that both are valid solutions. 30

31 Solving an Equation Involving Absolute Value Example 14 Solve x2-3x = -4x + 6 To solve this algebraically, consider the the expression inside the absolute value symbols can be positive or negative. This results in two separate equations, each of which must be solved. 31

32 Solving an Equation Involving Absolute Value Solution: Solve x2-3x = -4x + 6 First equation x2-3x = -4x + 6 x2 + x - 6 = 0 (x + 3)(x - 2) = 0 x+3=0 x = -3 x-2=0 x=2 32

33 Solving an Equation Involving Absolute Value Solution: Solve x2-3x = -4x + 6 Second equation -(x2-3x) = -4x + 6 x2-7x + 6 = 0 (x + 6)(x - 1) = 0 x-6=0 x=6 x-1=0 x=1 Now check to see if each solution is extraneous. 33

34 Solving an Equation Involving Absolute Value Solution: Solve x2-3x = -4x + 6 After checking each solution and observing the graph, you can see the x = -3 and x = 1 and the only two solutions. 34

35 True or False Justify your answer. (You can draw a graph that proves or disproves each question) 1. Two linear equations can have either one point of intersection of no points of intersection. 2. An equation can never have more than one extraneous solution 35

36 Exploration Given that a and b are nonzero real numbers, determine the solutions of the equations. 1. ax2 + bx = 0 2. ax2 - ax = 0 Hint: your solutions may be in the form of a and b 36

37 Exploration Solution: 1. ax2 + bx = 0 x(ax + b) = 0 Factor out the GCF either x=0 or ax + b = 0 ax = -b x = -b/a So your solutions are x = 0 and x = -b/a 37

38 Exploration Solution: 2. ax2 - ax = 0 ax(x - 1) = 0 Factor out the GCF either ax = 0 But we knew a is a nonzero number x=0 or x-1=0 So your solutions are x = 0 and x = 1 38

39 Homework B.3 Homework Due next class period 39

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