Solving Quadratic and Other Polynomial Equations

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Walter Thornton
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1 Section 4.3 Solving Quadratic and Other Polynomial Equations TERMINOLOGY 4.3 Previously Used: This is a review of terms with which you should already be familiar. Formula New Terms to Learn: Discriminant Your definition Formal definition Example Quadratic Formula Your definition Formal definition Example Zero Product Rule Your definition Formal definition Example READING ASSIGNMENT 4.3 Sections 6.6 and

2 Chapter 4 Factoring to Solve Polynomial Problems READING AND SELF-DISCOVERY QUESTIONS 4.3. What is the Quadratic Formula. What is the discriminant of a quadratic equation 3. How many solutions can a quadratic equation have 4. What is the Zero Product Rule 5. What is the first thing you do in solving a quadratic equation either by using the Quadratic Formula or by factoring KEY CONCEPTS 4.3 When dividing a polynomial by a polynomial where neither polynomial is a monomial, you first set-up the division in a form that uses the Division Algorithm. The quotient will be a constant or a polynomial with a remainder that can be presented in two ways. The Zero Product Rule is used in solving polynomial equations (including quadratic equations) as well as other types of equations. It states that the product of two numbers can be zero if and only if one of the factors in the product is zero or both of the factors are zero, that is, a b = 0 if and only if a = 0 or b = 0 or both a and b are 0. 70

3 TECHNIQUE 4.3 SOLVING POLYNOMIAL EQUATIONS OF DEGREE HIGHER THAN TWO BY FACTORING Some polynomial equations of degree higher than two can be solved by factoring using the Zero Product Rule. Limitation/Caution: Although formulas exist for solving polynomials of degree three and four (but not degree more than four), they are beyond the scope of this book. Most polynomial equations (including quadratic polynomials) cannot be solved by factoring because most polynomials cannot be factored with real numbers. Example : Solve = 5a for a. Step : Arrange the non-zero terms of the quadratic equation on one side in descending order so that it is in the standard form ax + bx + c = 0 Step : Using the Methodology for Factoring a Trinomial or the Technique for Factoring Polynomials of Higher Degree (Section 4.), factor fully. = 5 a 5a = 0 ( + )( a ) = 0 Step 3: Set each factor equal to zero and solve. + = 0 or a = 0 = or a = Step 4: Present the solution(s) in simplified form. a = or a = 3 Step 5: Validate by substituting solutions into the original equation. a = 3 3 = = = = 3 3 a = 3 = 5 = 0 0 = 0 7

4 Chapter 4 Factoring to Solve Polynomial Problems MODEL 4.3 Solve for x: x 3 + x x = 0 Step Place in standard form Already done. Step Factor + = 0 3 x x x ( ) 0 x x + x = x( x + 4)( x 3) = 0 Step 3 Set factors equal to zero and solve x = 0 x + 4 = 0 x 3 = 0 x = 0, x = 4, or x = 3 Step 4 Present solution(s) x = 0, x = 4, or x = 3 Step 5 Validate x = 0 x = 3 3 (0) + (0) (0) = = 0 x = 4 3 (3) + (3) (3) = = = 0 3 ( 4) + ( 4) ( 4) = = = 0 TECHNIQUE 4.3 SOLVING QUADRATIC EQUATIONS USING THE QUADRATIC FORMULA You use the Quadratic Formula to solve any quadratic equation by writing the equation in a standard form and then identifying and substituting the coefficients of the equation. Limitation/Caution: The technique often requires that you simplify a radical expression. Example: Solve x 4 5 = 6x for x Step : Arrange the non-zero terms of the quadratic equation on one side in descending order so that it is in the standard form ax + bx + c = 0 x 5 = 6 x x 6x 5 = 0 Step : Write down the values associated with a, b, and c. a =, b = 6 and c = 5 7

5 Step 3: Use the quadratic formula: ± x = b b 4ac a b ± b 4ac x = a ± x = () ( 6) ( 6) 4()( 5) 6 ± ± 76 6 ± 9 x = = = (3 + 9) 3± 9 x = = 4 Step 4: Present the solution(s) in simplified form. Step 5: Validate. 3 ± 9 x = OR x = 3.68, x = 0.68 x 5 = 6 x 3.68 x (3.68) 5 6(3.68) x 0.37 ( 0.68) 5 6( 0.68) CRITICAL THINKING QUESTIONS 4.3. What methods exist for solving a quadratic equation. Why are there at most two solutions to a quadratic equation 3. What does the symbol ± mean 73

6 Chapter 4 Factoring to Solve Polynomial Problems 4. If a polynomial p(x) factors into three linear factors, how many solutions will the equation p(x) = 0 have 5. Why do both the Quadratic Formula method and the factoring method for solving quadratic equations require that a zero appear on one side of the equation DEMONSTRATE YOUR UNDERSTANDING 4.3 Solve each of the following, using the method indicated.. 3x 6x = (Quadratic Formula). 3x 6x = (factoring) 3. a 5a 8 = + (Quadratic Formula) 74

7 4. a 7a + 0 = 0 (factoring) 4 IDENTIFY AND CORRECT THE ERRORS 4.3 In the second column, identify the error you find in each of the following worked solutions and describe the error made. Solve the problem correctly in the third column. Problem Describe Error Correct Process. Solve for a: + 4a = 5 Worked Solution (What is wrong here) Wrong process: Did not put the equation in standard form before determining the values of a, b, and c. 4a 5 a = 3, b = 4, c = 5 a = a = a = + = ± a b b 4ac ± a = (3) 4 4 4(3)(5) 4 ± 6 ± i 3. Solve for x: x( x )( x ) 3 + = 0 Wrong process: Did not take all factors into account. Worked Solution (What is wrong here) ( )( ) x 3x x + = 0 3x = 0 or x + = 0 x = or x = 3 75

8 Chapter 4 Factoring to Solve Polynomial Problems Problem Describe Error Correct Process 3. Solve for y: y y = 5 Worked Solution (What is wrong here) Wrong process: Did not put the equation in standard form with a zero on one side. y( y ) = 5 y = 0 or y = 0 y = 0 or y = 76

Solving Systems of Linear Equations

Section 2.3 Solving Systems of Linear Equations TERMINOLOGY 2.3 Previously Used: Equivalent Equations Literal Equation Properties of Equations Substitution Principle Prerequisite Terms: Coordinate Axes