Section The Quadratic Formula - Part I

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1 Math 17 - Section Page 1 Section The Quadratic Formula - Part I I. Derivation A. Remember that the standard form of a quadratic equation is: Ax + Bx + C = 0 B. We are now going to derive the quadratic formula (one of my all-time favorite math tasks!) by completing the square. So sit back and enjoy! C. The Derivation 1. First, we need to get the constant ( C ) by itself, so we subtract C from both sides to get: Ax + B -C. Now we need to get the coefficient of x to be 1, so we will divide everything by A to get: B C x + A A 3. Now take half of the coefficient of the linear term, square it, and add it to both sides to get: B B B C X + x + = A 4A 4A A 4. Now write the left side as a perfect square; find the LCD on the right to get: B B 4AC x + = A 4A. Now use the Square Root Property to get: 6. Subtract B A from both sides to get: B B 4AC x + = ± A 4A B B 4AC x + = ± A A B ± B 4AC A 7. This is the Quadratic Formula! II. Using the Formula A. Procedure 1. Write the equation in Standard Form (= 0).. Find A, B, & C. 3. Substitute into the formula. 4. Simplify the result. B. Examples Solve each of the following using the Quadratic Formula. 1. x + 1 = 10x First, add 10 x to both sides to put the equation in standard form: x + 10x + 1 = 0 Now identify A, B, & C: A =, B = 10, C = 1 Copyright 010 by John Fetcho. All rights reserved.

2 Math 17 - Section Page Now substitute to get: (10) (10) 4()(1) ± () 10 ± ± ± ± 4 = = = Factoring out the GCF in the numerator, we get: ( ± ) 10 We can now reduce to get: ±. x x + 61 = 0 This is already in Standard Form, so we need to identify A, B, & C: A = 1, B = 1, C = 61 Substituting into the formula, we get: ± (1) ( 1) ( 1) 4(1)(61) 1± ± 43 1± ± i 81 3 = = = 1± 9i 3 3. Now you try one: 3x 4x + = 0 ±i x + 3x 0 = 0 Already in Standard Form, so we identify A, B, and C: A =, B = 3, C = 0 Substituting into the formula, we get: (3) ± (3) () 4()( 0) 3 ± ± ± 13 = = Since the radical disappeared, we have more work to do! We have to actually find each of the values of x = = = = Copyright 010 by John Fetcho. All rights reserved.

3 Math 17 - Section Page 3 x= 4,. Now you try one: x + x 3 = 0 3, 1 III. Applications A. Remember that with each application, to get full credit, you need to include: 1. A definition for the variable.. An equation. 3. The work. 4. Answer the question with a complete sentence. B. Example - Solve the following. 1. When the shot is released at an angle of 6 o, its path can be modeled by the quadratic function f(x) = 0.04x +.1x in which x is the shot's horizontal distance, in feet, and f(x) is its height, in feet. This function is shown by one of the graphs, (a) or (b), in the figure at the left-hand column of page 748. Use the function to determine the shot's maximum distance. Use a calculator and round to the nearest tenth of a foot. Which graph, (a) or (b), shows the shot's path? (page 764, #76) First, the problem has defined our variables for us, but I'm going to restate it: Let The shot's horizontal distance, in feet Let f(x) = The shot's height, in feet The maximum distance is found when the shot hits the ground, so what do we set f(x) equal to? 0 = 0.04x +.1x Put in Standard Form by multiplying by 1. 0 = 0.04x.1x 6.1 Solve using the Quadratic Formula A = 0.04, B =.1, C = 6.1 ( ) (.1) ±.1 4(0.04)( 6.1).1± = (0.04) ±.386.1±.3 = Simplifying each piece, we get: = = OOPS! The shot lands about. feet from the shot putter. IV. Finding the equation from the roots. A. Suppose that we are asked to solve the quadratic equation x x + 4 = 0. Before we knew about the Quadratic Formula, we had to factor this. So let's solve this one by factoring: x x + 4 = 0 (x 4)(x 1) = 0 Factoring by Trial and Error x 4 = 0 x 1 = 0 Zero Product Property 4 1 Copyright 010 by John Fetcho. All rights reserved.

4 Math 17 - Section Page 4 B. What we are going to do now is start with the answer and generate the question (a la Jeopardy) as a quadratic equation. C. Examples - Write a quadratic equation in standard form with the given roots. 1. {, 6} What this means is: 6 We need to get each equation equal to 0. x + = 0 x 6 = 0 Now write the factors as a product. (x + )(x 6) = 0 Now multiply: x 4x 1 = 0 It should be noted that this is not the only possible "question". We could multiply the equation by any non-zero constant and still have the same solution set.. 1, 6 3 We begin by setting x equal to each root: In each case, multiply by the LCD to clear fractions Get each equation equal to 0. 6x + = 0 3x 1 = 0 Now write the product. (6x + )(3x 1) = 0 Multiply it out. 18x + 9x = 0 3. Now you try one: { 6i, 6i} x + 36 = 0 4. {1 + 3, 1 3 } Set x equal to each root: Get each equation equal to 0. x 1 3 = 0 x = 0 Write the product. [(x 1) 3 ][(x 1) + 3 ] = 0 Now recall that (a + b)(a b) = a b. This is really what we have, with (x 1) = a and 3 = b. So when we multiply, we get: (x 1) ( 3 ) = 0 Remember (x 1) = (x 1)(x 1). x x = 0 Combine like terms. x x = 0. Now you try one: {1 +, 1 } x x 1 = 0 Copyright 010 by John Fetcho. All rights reserved.

5 Math 17 - Section Page II. The Discriminant A. The discriminant is the portion of the quadratic formula inside the square root. In other words, the discriminant is B 4AC B. The discriminant tells us the type of solutions that a quadratic equation will have. Sign of the Discriminant Types of Solutions Negative (< 0) Two imaginary, conjugate roots. Equal to 0 (= 0) One real root (multiplicity ). Positive (> 0) Two real, unequal roots. C. If B 4AC is positive, our roots will be rational roots if B - 4AC is a perfect square. If B 4AC is not a perfect square, then the roots will be irrational. D. Examples - Describe the types of solutions. DO NOT SOLVE. 1. x x 3 = 0 We begin by finding A, B, and C: A = 1, B = 1, C = 3 Substituting these into the discriminant, we get: B 4AC = ( 1) 4(1)( 3) = = 13 > 0 The equation will have two real, irrational roots.. x x + = 0 Finding A, B, and C: A =, B = 1, C = Substituting these into the discriminant, we get: B 4AC = ( 1) 4()() = 1 16 = 1 < 0 The equation has two imaginary, conjugate roots. 3. Now you try one: 9x 4x + 16 = 0 The equation has one real root of multiplicity. Copyright 010 by John Fetcho. 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,