Algebraically and Graphically Solving Equations B Solvin… · Solving Equations Algebraically and ... Finding Solutions of an Equation Graphically ... (2(1) + 7) - (1) = 2 √(9)

B.3Solving Equations Algebraically and

Graphically

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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.

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Solving an Equation Involving Fractions

Example 1

Solve: x + 3x = 23 4

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Solving an Equation Involving Fractions

Example 1

Solve x + 3x = 23 4

Multiply each term by the LCD of 12 (12) x + (12) 3x = (12) 2

3 4Divide out and multiply 4x + 9x = 24 Combine like terms 13x = 24Divide each side by 13 x = 24

13Study Tip: after solving an equation, check each solution in the original equation.

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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?

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An Equation with an Extraneous Solution

Example 2Solve:

1 = 3 - 6x x - 2 x + 2 x2 - 4

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Example 2Algebraic Solution

1 = 3 - 6x x - 2 x + 2 x2 - 4

The LCD is x2 - 4 = (x + 2)(x - 2)(x + 2)(x - 2) 1 = (x + 2)(x - 2) 3 - (x + 2)(x - 2) 6x

x - 2 x + 2 x2 - 4

(x + 2) = 3(x - 2) - 6x x + 2 = 3x - 6 - 6x 4x = -8

x = -2

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.

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Example 2Graphical Solution:Use a graphing calculator to graph the left and right sides of the equation.y1 = 1 y2 = 3 - 6x

x - 2 x + 2 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

Intercepts and Solutions

Definition of Intercepts1. 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.

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Finding x- and y-Intercepts

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

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Finding x- and y-Intercepts

Example 3Solution2x + 3y = 5To 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

Finding Solutions Graphically

Graphical Approximations of Solutions of an Equation1. 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.

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Finding Solutions of an Equation Graphically

Example 4

Use a graphing calculator to approximate the solutions of 2x3 - 3x + 2 = 0

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Finding Solutions of an Equation Graphically

Solution:2x3 - 3x + 2 = 0Graph the function y = 2x3 - 3x + 2

You can see that there is one x-intercept between -2 and -1, at approximately -1.5.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 ≅ -1.476 Check this approximation in your calculator. 14

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.

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Finding Points of Intersection

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

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Algebraic Solution:Solve each equation for y2x - 3y = -2 -3y = -2x - 2

y = (⅔)x + ⅔Next, 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 20 = 10x

2 = x

4x - y = 6 4x = 6 + y 4x - 6 = y

When x = 2, the y-value of each of the original equations is 2. So, the point of intersection is (2, 2).

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

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Solving an Equation of Quadratic Type

Example 8Solve 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?

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Example 8Solve 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 Let u = x2

(u - 1)(u - 2) = 0 Factor(x2 - 1)(x2 - 2) = 0 Plug x2 back in

(x + 1)(x - 1)(x2 - 2) = 0x + 1 = 0 x = -1x - 1 = 0 x = 1x2 - 2 = 0 x = +/- √2

Check all four solutions algebraically 20


Example 8Solve x4 - 3x2 + 2 = 0 After checking each solution, you see that they are all valid.

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Solving an Equation Involving a Radical

Example 10Solve √(2x + 7) - x = 2

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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.

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Solution:√(2x + 7) - x = 2 √(2(-3) + 7) - (-3) =? 2 √(1) + 3 =? 2

1 + 3 =? 2 4 ≠ 2 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.

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Graphical Solution:√(2x + 7) - x = 2First rewrite the equation as √(2x + 7) - x - 2 = 0Now graph y = √(2x + 7) - x - 2Notice that the domain is x ≥ - 7/2 because the expression under the radical cannot be negative.

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Solving an Equation Involving Two Radicals

Example 11√(2x + 6) - √(x + 4) = 1

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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 = 1 + 2√(x + 4) + (x + 4) x + 1 = 2√(x + 4) Isolate 2√(x + 4) (x + 1)2 = [2√(x + 4)]2 Square each sidex2 + 2x + 1 = 4(x + 4)x2 + 2x + 1 = 4x + 16 Write in general formx2 - 2x - 15 = 0 Factor

(x - 5)(x + 3) = 0, so x = 5 and x = -3Check to see if either are extraneous solutions.

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Solution:√(2x + 6) - √(x + 4) = 1After checking each solution, you see that the only valid solution is x = 5

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Solving an Equation with Rational Exponents

Example 12Solve (x + 1)⅔ = 4

Hint: Rewrite as radical.

Don’t forget to check for extraneous solutions!

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Solving an Equation with Rational Exponents

Example 12Solve (x + 1)⅔ = 4

∛(x + 1)2 = 4 Rewrite in Radical form (∛(x + 1)2)3 = 43 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.

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Solving an Equation Involving Absolute Value

Example 14Solve | 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.

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Solution:Solve | x2 - 3x | = -4x + 6First equation

x2 - 3x = -4x + 6 x2 + x - 6 = 0

(x + 3)(x - 2) = 0 x + 3 = 0 x = -3 x - 2 = 0 x = 2

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Solution:Solve | x2 - 3x | = -4x + 6Second 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.

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Solution:Solve | x2 - 3x | = -4x + 6After checking each solution and observing the graph, you can see the x = -3 and x = 1 and the only two solutions.

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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

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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

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Exploration

Solution:1. ax2 + bx = 0 x(ax + b) = 0 Factor out the GCF

either x = 0orax + b = 0ax = -bx = -b/a

So your solutions are x = 0 and x = -b/a37

Exploration

Solution:2. ax2 - ax = 0 ax(x - 1) = 0 Factor out the GCF

either ax = 0 But we knew a is a nonzero numberx = 0orx - 1 = 0

So your solutions are x = 0 and x = 1

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Homework

➢ B.3 Homework➢ Due next class period

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Algebraically and Graphically Solving Equations B Solvin… · Solving Equations Algebraically and ... Finding Solutions of an Equation Graphically ... (2(1) + 7) - (1) = 2 √(9)

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