Mth 111 Topics. Chapter 1 Overview

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2 Mth 111 Topics Chapter 1 Overview Section 1.1 Solving Linear Equations Solving Linear Equations involving distributive property and foil Solving literal equations Applications average, mixture, uniform motion, work Section 1.2 Solving quadratics: o by factoring and zero product principle; o square root; o completing the square; o quadratic formula; Discriminant Applications: square box, tossing an object 2

3 Forming Equivalent Equations An equation can be rewritten into an equivalent form by: 1. Simplify either side 2. Using the Golden Rule of Algebra Do unto one side as thou hath done unto the other 3. Interchanging sides 3x = 7y equivalent to 3x + 5 = 7y + 5 Linear Equations : A linear equation in a single variable x is an equation that can be written in the form ax + b = 0 where a and b are real numbers, with a not 0. Solve 3 2(x 4) = 1 ( 2x + 5) 3

4 Solve (2x 2) (x + 3) = 2 (x 1) (x + 1) 2x 2 + 6x 2x 6 = 2 (x 2 +1x 1x 1) 2x 2 + 4x 6 = 2x 2 2 4x 6 = 2 4x = 4 x = 1 Literals Equations : formulas with several unknowns = a 4x, solve for x 2. 8 = ax + 3x, solve for x 3. P = S dt, solve for t 4

5 Word Problems 1. What do you need to know to solve the problem? Write this down in English. 2. Assign numbers to the known parts. Assign a letter to the unknown parts 3. Translate this into an algebraic equation or inequality. 4. Solve. 5. Make sure that your solution answers the original question Useful Formulas Business: Revenue = Profit Cost Simple Interest : I = Prt Mixtures: (1st % Amt) + (2nd % Amt) = Final % Amt Rate problems: Distance = rate time Work Problems: (part done by A ) + (part done by B) = 1 whole job Averages: (X 1 + X X N ) / N 5

6 Applications: Averages: (X 1 + X X N ) / N The average on your four tests is Your test scores are 92, 86, and 98 with one test score gone missing. Find it! Simple Interest : I = Prt You invest part of a $10,000 bonus in a 2% simple interest account and the remainder of the money at 5% simple interest. Together the investments earn $400 per year. find the amount at each rate. let x be amount in 2% let x be amount in 5% t = 1 I 2% + I 5% = 400 6

7 Rate problems : Distance = rate time Example : Two boats traveling the same direction leave a harbor at noon. After 3 hours they are 60 miles apart. If one boat travels twice as fast as the other, find the average rate of each boat. First we need to picture the situation. Let A be the slower boat and B be the faster one. Let x mph be the average rate of boat A. The average rate of boat B is 2x mph. Since they are 60 miles apart after 3 hours, the difference in the distance traveled by the faster boat B and the slower boat A is 60 miles. distance traveled by B distance traveled by A = 60 average rate of B time average rate of A time = 60 2x 3 x 3 = 60 6x 3x = 60 3x = 60, x = 20 2x = 40 The average rates are 20 mph and 40 mph 7

8 Mixtures : (1st % Amt) + (2nd % Amt) = Final % Amt How many gallons of cream containing 25% butter fat and milk containing 3.5% butter fat must be mixed to obtain 50 gallons of cream containing 12.5% butter fat? Let x = # gal cream Let 50 x = # gal milk 0.25x = # gal of butter fat in cream x = # gal of butter fat in milk = # gal of butterfat in mixture 0.25x x = x = rounded In the chemistry lab you have 20 oz of a 10% acid solution. How much 60% acid solution should you add to get a 35% acid solution? 8

9 Work Problems: (part done by A ) + (part done by B) = 1 whole job Ron, Mike, and Tim are going to paint a house together. Ron can paint one side of the house in 4 hours. To paint an equal area, Mike takes only 3 hours and Tim 2 hours. If the men work together, how long will it take them to paint one side of the house? Let t be time needed to paint the side. (1/4)t + (1/3)t + (1/2)t = 1 Section 1.2 Quadratic Equations and Their Applications * Solve by factoring * Solve by taking the square root * Solve by completing the square * Solve by using the Quadratic Formula * The discriminant of a Quadratic Equations Applications 9

10 A quadratic equation is an equation that can be written in the following standard form: ax² + bx + c = 0, where a,b,c are real numbers and a 0 The Zero Product Principle If A and B are algebraic expressions such that AB = 0, then A = 0 or B = 0 1.2: Factoring and using Zero Product Principle Solve by factoring... x 2 + x = 12 Solve by factoring... 2x 2 5x = 12 10

11 1.2: Square Root Procedure If x 2 = c, then x = c or c Solve by using the square root method. 2x 2 = 48 Solve by using the square root method. (x + 2) 2 = 36 Geometrically 11

12 1.2: Completing the Square Solve by completing the square x 2 + 6x 8 = 0 x 2 + 6x + = 8 + x 2 + 6x + 9 = ( x + 3 ) 2 = : Completing the Square Solve by "completing the square" x 2 2x 5 = 10 2 x 2 + 4x 4 = 0 12

13 a 0 Solve by using the quadratic equation : 12x 2 x 6 = 0 x 2 = 2x 2 1.2: Discriminant and solutions to the quadratic equation The equation ax 2 + bx + c = 0, with real coefficients and a 0, has as its discriminant b 2 4ac If b 2 4ac > 0 then two distinct real solutions. If b 2 4ac = 0 then one real solution. If b 2 4ac < 0 then two distinct nonreal complex solutions. The solutions are conjugates of each other 13

14 Determine discriminant and state the number of solutions a) 4x 2 5x + 3 = 0 b) x 2 + 2x 15 = 0 c) 4x x + 9 = 0 Now graphically 14