Solving Equations of Degree 1 (Linear Equations):

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1 Solving Equations of Degree 1 (Linear Equations): Definition: The equation a b0 is called an equation of degree 1. Eample: Solve for : 0. 0, Or. Dividing both sides by : giving. Easy! Check the answer! Substitute back into the original equation., 0. MATH 110 Lecture 1 of 0 Ronald Brent 016 All rights reserved.

3 Word Problems Eample: A piece of wire 76-in. long is bent into the shape of a rectangle that is -in. longer than it is wide. What is the area of the rectangle? + Since the perimeter is 76 inches we have ( ) 76. or 76 which is equivalent to 76, or 7, which has solution 18 inches. The dimensions of the rectangle are 18 by 0 inches, so the area is 60 sq. in. MATH 110 Lecture of 0 Ronald Brent 016 All rights reserved.

4 Eample: Two thousand dollars was invested, partly at % annual simple interest and partly at 7%. The total interest earned in the first year was $10. How much was invested at 7%? Let be the amount of money, in dollars, invested at 7%. The remaining money invested at % is 000. The amount of interest earned is (000 ) Multiplying by 100 gives 7 (000 ) 1,000 which reduces to 7 10,000 1,000 or 000, and hence 100 dollars is the desired answer. MATH 110 Lecture of 0 Ronald Brent 016 All rights reserved.

5 Solving Equations of Degree (Quadratic Equations): Definition: The equation a bc0 is called an equation of degree. Quadratic Formula: b b ac MATH 110 Lecture of 0 Ronald Brent 016 All rights reserved. a Eample: 0. Here a = 1, b =, and c =. So, according to the quadratic formula ( ) ( ) , or. 1 When the "" sign is used, we get the solution 6. When the "" sign is used, we get the solution 1. So there are two solutions: = and = 1.

6 Three Cases for a Quadratic Equation The number b ac is called the discriminant, since it allows you to discriminate among the following three cases that indicate what types of roots you will have. 1) b ac 0: In this case, the quadratic equation has two real and distinct roots that is, two roots that are different numbers. ) b ac 0: Here, the square root in the solution disappears, leaving one real repeated root. ) b ac 0: In this last case, the roots are comple numbers. MATH 110 Lecture 6 of 0 Ronald Brent 016 All rights reserved.

7 MATH 110 Lecture 7 of 0 Ronald Brent 016 All rights reserved. Eamples: a) 8 6 Here the discriminant is ) ( () (6) 6 The roots are = 1, and = b) 9 Here the discriminant is ) 9 ( (1) () The roots are repeated with c) Here the discriminant is 0. 0 () () () i The roots are i 1

8 Word Problem Eample: A football field is 100-yd long from goal line to goal line and 160-ft wide. If a player ran diagonally across the field from one goal line to the other goal line, how far did he or she run? 100 yd = 00 ft 160 ft Using the Pythagorean Theorem , and so,600 90,000 11, 600. Taking the square root of both sides gives 11, The only physically reasonable answer is = 0 ft. MATH 110 Lecture 8 of 0 Ronald Brent 016 All rights reserved.

9 Eample: A rectangle 6 by 10 is to be reduced in size by cutting two strips of equal width, one from the length, the other from the width, so that the area is reduced by 1 sq in. How wide should the strips be? s.. 10 So, The original area is 60 sq. in., hence ( 10 )(6 ) 601. This gives the equation 60 16, which can be rewritten as Using the quadratic formula, we obtain 16 (16) (1)(1) The only reasonable solution here is = 1 inch , 1 MATH 110 Lecture 9 of 0 Ronald Brent 016 All rights reserved.

10 Solving Other Types of Equations Some of the methods that we have used to solve equations of degree 1 and can also be used to solve other kinds of equations. There are four basic tools in our equation solving toolbo: 1) Peeling the Onion; for any equation where the variable that we wish to solve for occurs in only one place, or any equation that can be put into that form. ) The Quadratic Formula; for equations that are of quadratic type, meaning that a well-chosen substitution can change the given equation into an ordinary quadratic equation. ) The Zero-Factor Property; which allows us to solve equations of the form f ( ) 0, by factoring f (), and setting each factor equal to zero. ) Rational Equations; For cases where the equations have fractions and polynomials in the denominator. MATH 110 Lecture 10 of 0 Ronald Brent 016 All rights reserved.

13 Equations Reducible to Quadratics: m m Any equation like a t bt c 0, can be reduced to a quadratic equation by letting m t. Eample: Solve for t t t 0. Let t, then This has solutions 1, and. So 1 or. With t, t, and so the solutions to the original problem are t 1 1, and t. Eample: Solve for y y y 0. Let y, then y, and the equation becomes 0. As before the solution here are 1 or. Now with y, y, and so y 1, and y. MATH 110 Lecture 1 of 0 Ronald Brent 016 All rights reserved.

14 Eample: Solve for y y y 0. Again we let y, then y, and the equation becomes 0. Using the quadratic formula now gives 1 or. Now with y, y, and so y 1 i, and y i. Eample: Solve for y y 7 y 0. Let y, then y, and the equation becomes 7 0. This has solutions Now with y 7 1, y, and so 7 1 y. MATH 110 Lecture 1 of 0 Ronald Brent 016 All rights reserved.

15 Zero-Factor Property If ab 0 then either a = 0, or b = 0, (Or both equal 0) Eample: Suppose ( 1)( ) 0 Then either ( 1) 0 or ( ) 0 Hence 1 or This etends to equations with more than one term. Eample: Suppose ( )( )( 1) 0 Then either ( ) 0 or ( ) 0 or ( 1) 0 And so or or MATH 110 Lecture 1 of 0 Ronald Brent 016 All rights reserved.

16 Eample: Suppose 1 0. Factoring gives ( )( ) 0 so either ( ) 0 or ( ) 0. In the first case and in the second case Eample: Suppose Factoring gives ( )( ) 0 so either ( ) 0 or ( ) 0. In the first case and in the second case i MATH 110 Lecture 16 of 0 Ronald Brent 016 All rights reserved.

17 Rational Equations: Eample: (1 ) Find the LCD for the given fractions and multiply the equation by that. The LCD is ( 1) or [ (1 1 1 )] 1 [ (1 )] 1 1 (1 ) or 1 MATH 110 Lecture 17 of 0 Ronald Brent 016 All rights reserved.

19 Eercises Find the solutions, both real and comple, for Eercises 10. 1) Solve for : 0. ) Solve for z: 1 z 1. ) Solve for : ) Solve for y: y y. ) Solve for : y z 0. 6)Solve for y: z y z ) 0 8) ) y y ) y 8y ) 1 0 1) 7 0 1) g g 0 1) ) y 0 16) 17) y 18. y ) ( )( 81) 0 19) z 8z ) 6 0 1) s 9s 0 0 ) 0 ) 1 ) Find all real solutions to 0. ) Find all solutions to 6) Find all solutions to (Hint: is a common denominator for the two algebraic fractions on the LHS.) ) Find all real solutions to ) Find all real solutions to ) Find all real solutions to. 0) Find all real solutions to. MATH 110 Lecture 19 of 0 Ronald Brent 016 All rights reserved.

20 Solutions: 1) = 8 ) z ) 0 7 ) 7 1 y ) z y 6) z 1 y 7) 1, 8) 9) 1 y, 1 10) y z 11) 1) 7 1) g, 1 1) 1 1, 1) y i 16) 1 17) y, 1 18), 1, 19) z 0), i s ) 16 ) 16 ) 1, ) 1 1), 6) No Solution 7) 8), 1 9), 1 0) MATH 110 Lecture 0 of 0 Ronald Brent 016 All rights reserved.

1.3 Algebraic Expressions

1.3 Algebraic Expressions A polynomial is an expression of the form: a n x n + a n 1 x n 1 +... + a 2 x 2 + a 1 x + a 0 The numbers a 1, a 2,..., a n are called coefficients. Each of the separate parts,