M141 - Chapter 1 Lecture Notes Page 1 of 27 Section 1.4 Solving Other Types of Equations Objectives: Given a radical equation, solve the equation and check the solution(s). Given an equation that can be reduced to quadratic form, solve equation. Given a rational equation, find all of the solutions. Given an application that results in a quadratic equation, find the solution. Solving Equations in General A reminder on what the big picture is when solving equations. some expression = another expression Goal is: find the of the variable that make the equation Example: 2x + 1 = x + 5 What value(s) of the variable x make this linear equation true?
M141 - Chapter 1 Lecture Notes Page 2 of 27 Solving Radical Equations Example: 4 x 1 5 x Technique: Example: Original equation: x = 3 solution is: x = 3 Square both sides: New equation: solution is: Notice that only ONE of the solutions of the new equation satisfies the original equation. Solutions of the original equation will always be a solutions of the new squared equation. of the must CHECK every solution of the new equation back in the original equation to eliminate the solutions
M141 - Chapter 1 Lecture Notes Page 3 of 27 Example: 4 x 1 5 x
M141 - Chapter 1 Lecture Notes Page 4 of 27 Next: solving an equation with two radicals in it: Example: 3x 1 2 x 1 Technique: isolate one of the radicals on either side: all other stuff = radical square both sides, just like before Still have a radical to get rid of, so need to repeat the procedure: isolate the radical again square both sides Check solution in original equation!
M141 - Chapter 1 Lecture Notes Page 5 of 27 Solve the equation. 1. 4n + 8 = n + 3 2. x 5 x 5
M141 - Chapter 1 Lecture Notes Page 6 of 27 Solving Rational Equations Rational equations: contain rational expressions, or expressions that are, where the variable appears in the of the expression. Method for Solving Rational Equations: Find the of all denominators. Multiply both sides of the equation by the (remember to multiply ALL terms). This will clear the equation of all fractions. Continue solving using techniques for a linear or quadratic equation. VERY IMPORTANT the solutions that you get for x are solutions. MUST check the possible solutions back in the original equation. If a possible solution causes a denominator to be in the original equation it is NOT a solution. Check the final solutions to verify that the equation is satisfied. Example: 4 2x 2 x 3 x 9 1 x 3
M141 - Chapter 1 Lecture Notes Page 7 of 27 Solve. 1. 2x 1 = 27 x 6 x+6 x 2 36 2. n 2 n 6 n 2 2n+12 n = n 6 2n
M141 - Chapter 1 Lecture Notes Page 8 of 27 Solving Equations that are Quadratic in Form Means: Equation is not quadratic, but Can make a variable substitution and into a quadratic equation of form: Uses a substitution method, where we will start by writing: Let u = Example: 4m + 8 m 21 = 0 Let u = u 2 = Substitute: Solve for u: Solve for m: Check solutions in original eqn:
M141 - Chapter 1 Lecture Notes Page 9 of 27 Summary of method: 1. Get: all stuff = 0 2. Make a substitution: Let u = 3. Substitute u and u 2 to create a quadratic equation. 4. Solve the quadratic equation for u. 5. Solve for the original variable. 6. Go back and check the possible solutions for the original variable in the original equation eliminate extraneous solutions. Solve: 1. (x 2 3) 4 x 2 3 12 = 0 Hint: look at it first, don t start by combining the constants!
M141 - Chapter 1 Lecture Notes Page 10 of 27 Solving Application Problems with Quadratic Equations Many of these problems start out as rational equations, then are transformed into quadratic equations. Techniques to solve a quadratic equation: 1. Factoring, then using the property 2. Using the property 3. Completing the 4. Using the formula Example: A civic club charters a bus for one day at a cost of $575. When two more people join the group, each person s cost decreases by $2. How many people were in the original group? (Note that the total cost of $575 remains the same even with the addition of two more people.) Original group New group Number of people Cost per person
M141 - Chapter 1 Lecture Notes Page 11 of 27 Problem: A small swimming pool can be filled by two hoses together in 1 hour and 12 minutes. The larger hose alone fills the pool in 1 hour less than the smaller hose. How long does it take the smaller hose to fill the pool? Note: To set up the equation for this problem, it is helpful to remember the Work Principle which you may have covered in Algebra. It always escapes my memory, so here it is: Also note that you need consistent time units either convert to straight hours, or to minutes.
M141 - Chapter 1 Lecture Notes Page 12 of 27 Section 1.5 Inequalities (Polynomial and Rational) Objectives: solve quadratic/polynomial inequalities solve rational inequalities Solving Quadratic Inequalities Example: f(x) = x 2 2x 8 a quadratic function Quadratic equation: x 2 2x 8 = 0 Solutions: Quadratic Inequalities: 1. x 2 2x 8 > 0 Means: for what x-values is 2. x 2 2x 8 0 Means: for what x-values is
M141 - Chapter 1 Lecture Notes Page 13 of 27 Fact: A polynomial will have a constant sign (either + or ) within each interval between its real zeros. Note that we are talking about the output value, or the y-value of the polynomial here. The two zeros divide the number line (x-axis) into intervals. In each interval, the function value is either. Using the Test-Point Method: Draw a number line, representing the x-axis Label the points where the zeros occur (find these by factoring, or using QF) Pick some x-value (ANY x-value) in that interval Calculate the corresponding function value (y-value) by plugging x into the function Determine if that y-value is positive or negative ALL y-values in that interval will be the same sign, either positive or negative f(x) = x 2 2x 8 = (x 4)(x + 2) Interval: Test point: f(x): Sign of f(x): Solution to: x 2 2x 8 > 0
M141 - Chapter 1 Lecture Notes Page 14 of 27 Example: 6x 2 < 6 + 5x Get all stuff on one side: Find the zeros: Use the Test-Point Method: f(x) = 6x 2 5x 6 Interval: Test point: f(x): Sign of f(x): 6x 2 5x 6 < 0
M141 - Chapter 1 Lecture Notes Page 15 of 27 Solving Polynomial Inequalities Example: 3x 3 12x > 0 Find: what x-values give a function value. From graph, zeros are: Solution is: Without graph, use the Test-Point Method: get all stuff on left-hand side inequality 0 on right hand side Find the zeros by factoring and/or using quadratic formula Draw a number line, representing the x-axis, and label the zeros Pick some x-value (ANY x-value) in that interval Calculate the corresponding function value by plugging x into the function Determine if that y-value is positive or negative ALL y-values in that interval will be the same sign, either positive or negative 3x 3 12x > 0 Interval: Test point: f(x): Sign of f(x): 3x 3 12x > 0
M141 - Chapter 1 Lecture Notes Page 16 of 27 Solve the polynomial inequality: 1. x 2 8x + 7 > 0 2. x 3 + 8x 2 + 15x > 0
M141 - Chapter 1 Lecture Notes Page 17 of 27 Solving Rational Inequalities Rational expressions: Example: rational function: Generically: Rational inequality in standard form: f(x) = 3 x x 5 Domain: Vertical asymptote: Horizontal asymptote: The number line can be split up into intervals based on: the of the function (where the graph crosses the x-axis), which is just like we saw above for the quadratics/polynomials, but also the Where is the function equal to 0?
M141 - Chapter 1 Lecture Notes Page 18 of 27 Fact: A rational expression Q P will have a constant sign (either + or ) within each interval determined by the real zeros of P and Q. Find the zeros of both the numerator and the denominator here, to divide the number line, BUT the zeros play different roles: At the zeros of P: the function value (y) will At the zeros of Q: the function value (y) is, therefore there will be a vertical asymptote. Method to use: put into standard form: find all zeros for P and Q (set P = 0 and solve, set Q = 0 and solve) plot zeros on number line choose test number in each interval and find sign (+/ ) write the solution to the inequality IMPORTANT The zeros of Q may be included in the solution set, even if the inequality is Division by zero is undefined. automatically put an on the number line at the Q zeros.
M141 - Chapter 1 Lecture Notes Page 19 of 27 3 x Example: 0 x 5 analytic solution: Set P = 0: 3 x = 0 when Set Q = 0: x + 5 = 0 when f(x) = 3 x x 5 Interval: Test point: f(x): Sign of f(x): Solution: How to put an inequality into standard form, if it isn t already: 3 x Example: 5 x 5
M141 - Chapter 1 Lecture Notes Page 20 of 27 Solve the rational inequality: 1. x x 2 0 2. 2x 3 x+3 1
M141 - Chapter 1 Lecture Notes Page 21 of 27 Section 1.6 Equations and Inequalities Involving Absolute Value Objectives: Solve equations and inequalities involving absolute value. Absolute value The absolute value of a number is its the number line. from the origin on Absolute value is always. distance = distance = 0 Example: x = 4 Solutions are: Either: x = or x = Equations with Absolute Value Example: x 1 = 4 Solutions are: Either: Check:
M141 - Chapter 1 Lecture Notes Page 22 of 27 Technique for Solving expression = a number From previous example, x 1 = 4, notice that you CANNOT simplify the problem by:. The goal is to: the absolute value term then the absolute value bars CAUTION!!! What if the problem was the following? Example: x 1 = 4 What is the solution? KEY point: ALWAYS check your solutions back in the original equation when working these absolute value problems. You might find that both solutions work, one solution works, or NEITHER works!!!
M141 - Chapter 1 Lecture Notes Page 23 of 27 Technique for Solving: expression = different expression In other words: u = v, where both u and v are algebraic expressions. This is ALSO the same as the first case: x = 4 To solve the equation: u = v Either: Example: 4x + 7 = x + 1
M141 - Chapter 1 Lecture Notes Page 24 of 27 Equations with Two Absolute Value Expressions In other words: u = v, where both u and v are algebraic expressions. This is ALSO the same as the first case: x = 4 If we had: x = 4 This simplifies to: To solve the equation: u = v Either: Inequalities with Absolute Value Example: x < 4 The solution to this will be an discrete number of two solutions to the equation. of values, as opposed to a The statement means: what values of x are from the origin. 4 units away Solution is: Example: x 4 Solution is:
M141 - Chapter 1 Lecture Notes Page 25 of 27 Example: x > 4 The solution to this will be an discrete number of two solutions to the equation. of values, as opposed to a The statement means: what values of x are from the origin. 4 units away Solution is: Example: x 4 Solution is:
M141 - Chapter 1 Lecture Notes Page 26 of 27 Special Cases of Inequalities: 1. expression > some negative number Example: 2x 1 > 5 Solution is: 2. expression < some negative number Example: 2x 1 < 5 Solution is: Example: 4+ 2 x 2 16 First isolate the absolute value: Next get rid of absolute value bars:
M141 - Chapter 1 Lecture Notes Page 27 of 27 Problems: 1. 2 2x 7 + 11 = 25 2. 2x 8 = x + 3 3x 2 3. 1 5