Hartfield College Algebra (Version 2015b - Thomas Hartfield) Unit ONE Page - 1 - of 26 Topic 0: Review Solving Quadratic Equations Three methods to solve equations of the form ax 2 bx c 0. 1. Factoring the expression and applying the Zero Product Property 2. Completing the square and applying the Property of Square Roots 3. Applying the Quadratic Formula Ex. 1 Solve by factoring and applying the Zero Product Property. 2 x 2x 48 0
Hartfield College Algebra (Version 2015b - Thomas Hartfield) Unit ONE Page - 2 - of 26 Ex. 2 Solve by completing the square and applying the Property of Square Roots. 2 x 6x 2 0 Ex. 3 Solve by applying the Quadratic Formula. 2 b b 4ac 2 x 2x 4x 1 0 2a
Hartfield College Algebra (Version 2015b - Thomas Hartfield) Unit ONE Page - 3 - of 26 Ex. 4 Solve by each of the previous two methods. 2 x 8x 4 0
Hartfield College Algebra (Version 2015b - Thomas Hartfield) Unit ONE Page - 4 - of 26 Topic 1: Solving Equations by Factoring Ex. 2 The first two equations can be solved after a GCF is factored out of the expression. 5 4 3 x 3x 10x 0 Ex. 1 x 5 2 64x
Hartfield College Algebra (Version 2015b - Thomas Hartfield) Unit ONE Page - 5 - of 26 The next three equations are described as quadratic-type when an appropriate substitution is applied. Ex. 3 2 2 2 4 7 4 6 0 x x Ex. 4 2 1 3 3 x 7x 12 0
Hartfield College Algebra (Version 2015b - Thomas Hartfield) Unit ONE Page - 6 - of 26 Ex. 5 4 2 x 9x 8 0
Hartfield College Algebra (Version 2015b - Thomas Hartfield) Unit ONE Page - 7 - of 26 The next two equations involve expressions that are factorable by the grouping method. Ex. 7 3 2 2x x 18x 9 0 Ex. 8 3 2 x 2x 3x 6 0
Hartfield College Algebra (Version 2015b - Thomas Hartfield) Unit ONE Page - 8 - of 26 Topic 2: Radical Equations Ex. 2 Radical equations have one or more radical expressions in the equation. To solve, isolate a radical on one side and then square both sides. Solve the resulting equations and check your solutions. x x 3 5 Ex. 1 2 x 3 1 13 Point of emphasis: Solving a radical equation by this method tends to create extraneous solutions so you must always check your results.
Hartfield College Algebra (Version 2015b - Thomas Hartfield) Unit ONE Page - 9 - of 26 Ex. 3 2x 6 x 3
Hartfield College Algebra (Version 2015b - Thomas Hartfield) Unit ONE Page - 10 - of 26 Topic 3: Rational Equations Rational equations have one or more rational expressions in the equation. To solve, find the least common denominator and multiply the LCD to both sides of the equation. Solve the resulting equation and check your results. Ex. 1 x x 1 1 2x 7 x 3
Hartfield College Algebra (Version 2015b - Thomas Hartfield) Unit ONE Page - 11 - of 26 Ex. 2 2x 2 4 1 2 x 7x 12 x 3 x 4
Hartfield College Algebra (Version 2015b - Thomas Hartfield) Unit ONE Page - 12 - of 26 Topic 4: Absolute Value Equations For the equation x c, there are three cases to consider: If c > 0, then the solutions can be found by x c. If c = 0, then the solutions can be found by x 0. If c < 0, then there are no solutions. As necessary, take steps to isolate the absolute value expression on one side before applying the previous rule. Ex. 2 4 x 1 6 14 3 2 Ex. 1 4x 5 9
Hartfield College Algebra (Version 2015b - Thomas Hartfield) Unit ONE Page - 13 - of 26 Ex. 3 x 1 7 2 1 3
Hartfield College Algebra (Version 2015b - Thomas Hartfield) Unit ONE Page - 14 - of 26 Topic 5: Linear & Compound Inequalities; Interval Notation Inequalities, like equations, are statements of relationships. Ex. 2 x 4 2 Equations state that two expressions are equal while inequalities state that expressions are not necessarily equal. Solving linear inequalities and solving linear equations require similar techniques with the familiar appendant that multiplying or dividing by a negative causes the inequality to reverse. Ex. 1 35 2x 1 Ex. 3 3x 18
Hartfield College Algebra (Version 2015b - Thomas Hartfield) Unit ONE Page - 15 - of 26 One significant difference is in the solution set. Equations typically have distinct, discrete values in their solution set. Inequalities, on the other hand, typically have continuous intervals of solutions. We graph inequalities on a number line so that we can see the solution set. To graph the solution set of an inequality: Mark the boundary value(s) of the solution interval. At each boundary value, note the inclusion or exclusion of the boundary in the solution set. Interval notation allows us to identify a continuous interval of values. We often associate interval notation with inequalities since the solution sets of inequalities are usually continuous intervals. Interval Notation ( ) parentheses boundary excluded or, or brackets [ ] boundary included If the boundary is included, use a closed dot If the boundary is excluded, use an open dot Shade appropriately using test values to identify the intervals. left end boundary or right end boundary or
Hartfield College Algebra (Version 2015b - Thomas Hartfield) Unit ONE Page - 16 - of 26 Sketch the graph of the inequality on the number line and write the corresponding interval of values Ex. 4a x 3 Symbols/terminology relevant to intervals & interval notation: Closed Interval: 2,7 both numeric boundaries included Open Interval: 4,3 both numeric boundaries excluded Ex. 4b x 1 Half-open Interval: 5, 1, 0,4 one end open, one end closed Infinite Interval:,4, unbounded on one or both ends 5,,, Ex. 5a 4 x 0 Ex. 5b 1 x 15 Union Symbol: <means or> each entire interval appropriate Intersection Symbol: <means and> only common (i.e. overlapping) interval appropriate
Hartfield College Algebra (Version 2015b - Thomas Hartfield) Unit ONE Page - 17 - of 26 Compound Inequalities (also known as Simultaneous or Three-Part Inequalities) Ex. 6 Present solutions graphically & 5 2x 1 13 Points for observation: It is often possible to solve a compound inequality be applying a common step to all three parts of the inequality. Certain compound inequalities must be rewritten as follows to be solved: A < B < C implies that A < B and B < C
Hartfield College Algebra (Version 2015b - Thomas Hartfield) Unit ONE Page - 18 - of 26 Topic 6: The less than case: Absolute Value Inequalities As with absolute value equations, with absolute value inequalities you want to isolate the absolute value expression. If x c, then c x c. (correspondingly true for ) Ex. 2 Present solutions graphically & Ex. 1 Present solutions graphically & 3x 2 5 5 3 x 2 13
Hartfield College Algebra (Version 2015b - Thomas Hartfield) Unit ONE Page - 19 - of 26 Ex. 3 Present solutions graphically & Ex. 4 Present solutions graphically & 2x 1 4 17 2 x 5
Hartfield College Algebra (Version 2015b - Thomas Hartfield) Unit ONE Page - 20 - of 26 The greater than case: If x c, thenx c or x c. (correspondingly true for ³) Ex. 6 Present solutions graphically & 5 2 x 3 13 Ex. 5 Present solutions graphically & 4x 5 7
Hartfield College Algebra (Version 2015b - Thomas Hartfield) Unit ONE Page - 21 - of 26 Exceptional scenarios: Ex. 8a Ex. 7a 2x 9 3 2x 9 3 Ex. 7b 2x 9 0 Ex. 8b 2x 9 0
Hartfield College Algebra (Version 2015b - Thomas Hartfield) Unit ONE Page - 22 - of 26 Topic 7: Polynomial Inequalities Polynomial inequalities are a type of nonlinear inequality which must be solved in a very different manner than linear inequalities. Whereas our approach to solving linear inequalities is relatively straight-forward, our technique for solving nonlinear inequalities attacks the problem sideways, in a manner of speaking. Process: 1. As necessary, rewrite so that one side is zero. 2. As necessary, factor the expression. 3. Set each factor equal to zero and solve to find boundary values for the solution intervals. 4. Draw a number line and mark the boundaries with an appropriate dot (closed if equal to, open if not). 5. Pick a test value from each interval formed by the boundaries and evaluate the expression using each one. Only the sign of the evaluation is relevant (all positive numbers are greater than zero, all negatives are less than zero.) 6. Identify the solution intervals. Ex. 1 Present solutions graphically & x x 2 1 4 0
Hartfield College Algebra (Version 2015b - Thomas Hartfield) Unit ONE Page - 23 - of 26 Ex. 2 Present solutions graphically & Ex. 3 Present solutions graphically & 3 2 x 2x 4x 8 3 2 x 5x 14x 0
Hartfield College Algebra (Version 2015b - Thomas Hartfield) Unit ONE Page - 24 - of 26 Topic 8: Rational Inequalities Rational inequalities are another type of nonlinear inequality which requires a circuitous approach to finding the solutions. If a rational inequality involves a rational expression compared to zero, the approach is almost exactly the same as a polynomial inequality. Both the numerator and denominator need to be factored (when appropriate) and the most significant difference is that any boundary created from the denominator must be open regardless of the inequality symbol (since setting the denominator equal to zero cannot produce a solution). Ex. 1 Present solutions graphically & 2x 3 0 2 x 2x When a rational inequality does not involve a comparison to zero, the complexity of the process is increased because it is necessary to rewrite the inequality so that one side is zero.
Hartfield College Algebra (Version 2015b - Thomas Hartfield) Unit ONE Page - 25 - of 26 Ex. 2 Present solutions graphically & 3 x 1 2 x 4
Hartfield College Algebra (Version 2015b - Thomas Hartfield) Unit ONE Page - 26 - of 26 Ex. 3 Present solutions graphically & x 4 1 2x 1