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Unit 5 Bellringer 11/12/14 1. 2. What is a variable? a letter used to represent a value or unknown quantity that can change or vary. What is a coefficient? a number being multiplied by a variable 3. What is a term? a number, a variable, or the product of a number and variable(s). How are quadratic expressions and quadratic equations alike? How are they different? Algebraic expressions are mathematical statements that include numbers, operations, and variables to represent a number or quantity. They do NOT have an = If you set an algebraic expression equal to something it becomes an algebraic equation. The left side of the = has the same value as the right side. Linear expressions are expressions where the highest power of the variable is the first power. Example: 2x+1 Today we are going to work with expressions and equations to the second power. Quadratic expressions are expressions where the highest power of the variable is the second power. Example: 3𝑥 2 − 5𝑥 + 4 Quadratic Expressions in standard form are written as 𝑎𝑥 2 + 𝑏𝑥 + 𝑐, where x is the variable and a, b, and c are constants. 𝑏 and 𝑐 can be any value, but 𝑎 can never equal zero Determine whether it is a quadratic expression. 6(x – 1) – x(3 – 2x) + 12. 6(x – 1) – x(3 – 2x) + 12 Original expression 6x – 6 – x(3 – 2x) + 12 Distribute 6 over x – 1. 6x – 6 – 3x + 2x2 + 12 Distribute –x over 3 – 2x. 3x + 6 + 2x2 2x2 + 3x + 6 Combine like terms: 6x and –3x; –6 and 12. Rearrange terms so the powers are in descending order. Determine whether it is a quadratic expression. Identify each term, coefficient, and constant. Classify the expression as a monomial, binomial, or trinomial. 1. 2𝑥2 + (3𝑥 + 1) + (3𝑥 + 2) 2. 4𝑥 − (5𝑥 + 8) 3. (𝑥 + 5) (2𝑥 − 7) Bellringer 11/13/14 1. Is 3𝑥 − 4 𝑥 + 2 + 𝑥(1 + 6𝑥) a Quadratic expression? Yes it is, Simplified: 6𝑥 2 − 8 2. What are the terms? The terms are 6𝑥 2 and -8 How are quadratic expressions and quadratic equations alike? How are they different? Determine whether it is a quadratic expression. Identify each term, coefficient, and constant. Classify the expression as a monomial, binomial, or trinomial. 1. 2𝑥2 + 3𝑥 + 1 + 3𝑥 + 2 = 2𝑥 2 + 6𝑥 + 3, yes it is a quadratic trinomial 2. 4𝑥 − 5𝑥 + 8 = −x − 8 3. 𝑥 + 5 2𝑥 − 7 = 2𝑥 2 + 3𝑥 − 35 A factor is one of two or more numbers or expressions that when multiplied produce a given product. Example: 3 and 4 are factors of 12, (3*4=12) In practice problem 3 from yesterday #3. 𝑥 + 5 2𝑥 − 7 simplified to 2𝑥 2 + 3𝑥 − 35 𝑥 + 5 𝑎𝑛𝑑 (2𝑥 − 7) are factors of 2𝑥 2 +3𝑥 − 35. When presented with a polynomial in factored form, such as example 3, multiply the factors to see if the polynomial in standard form is a quadratic. • Sometimes it is easier to work with the factored form to determine the values that make an expression negative or equal to 0. The length of each side of a square is increased by 2 centimeters. How does the perimeter change? How does the area change? Define what you know, don’t know and what you need to find out. We know a square has 4 equal sides and we are increasing each side by 2 We don’t know the lengths of the sides of the original square We need to find the Perimeter(add the lengths of all the sides) and Area (multiply length times width) and how they change between the original square and the increased square. We don’t know the length of the sides of the original square so we represent it with a variable, x X=the length of 1 side of a square Perimeter of this square is: 𝑥 + 𝑥 + 𝑥 + 𝑥 = 4𝑥 Area of this square is: 𝑥 ∗ 𝑥 = 𝑥2 x x x x The increased square’s sides are increased by 2 Does this mean we add, subtract, multiply, or divide by 2? Increase means we add, so our new sides equal x+2 Perimeter of this square is: 𝑥 + 2 + 𝑥 + 2 + 𝑥 + 2 + 𝑥 + 2 = 4𝑥 + 8 Area of this square is: 𝑥 + 2 ∗ 𝑥 + 2 = 𝑥 2 + 4𝑥 + 4 x+2 x+2 x+2 x+2 x x x x Perimeter: 4x Area: 𝑥 2 To see how the Perimeter and Area change we take the increased square Perimeter and Area and subtract the original square Perimeter and Area Change in Perimeter: (4x+8)-(4x)=8 Change in Area: ( 𝑥 2 +4𝑥 + 4)-(𝑥 2 )=4x+4 x+2 x+2 x+2 x+2 Perimeter: 4x+8 Area: 𝑥 2 +4𝑥 + 4 The length of each side of a pentagon is increased by 2 centimeters. All of the pentagon’s 5 sides are equal. How does the perimeter change? Define what you know, don’t know and what you need to find out. Bellringer 11/14/14 1. Identify the Terms, Coefficients, 2 and Constants in 2𝑥 + 4𝑥 Terms: 2𝑥 2 , and 4𝑥; Coefficients: 2, 4; Constants: 0 2. What are Factors? 2 or more expressions multiplied to get a certain outcome 3. Are (𝑥) and (2𝑥 + 4) factors of the expression in Question 1? Yes, when you multiply the factors you will get 2𝑥 2 + 4𝑥 How does changing the value of a coefficient, constant, or variable in an expression change the value of the expression? What do we know, not know, and need to find We know: It is a triangle The sides of the triangle are 8x, 10x, and 11x-2 X is the length in feet We do NOT know: What to use for x We need to find: … 1) Find an expression for the perimeter of the park. Set up an equation using the given expressions. Then combine like terms to find an expression to represent the perimeter, P. P = 8x + 10x + (11x – 2) 10x 8x P = 8x + 10x + 11x – 2 P = 29x – 2 11x-2 2) What long? is the perimeter of the park if a fence panel is 4 feet Substitute 4 for x in the equation for perimeter. P = 29(4) – 2 = 116 – 2 = 114 If a fence panel is 4 feet long, the perimeter of the park is 114 feet. 3) How does the perimeter change if the length of each fence panel increases to 5 feet? Substitute 5 for x in the equation for perimeter and compare the result to the previous question’s answer. P = 29(5) – 2 = 145 – 2 = 143 If the length of each fence panel increases to 5 feet, the perimeter of the park increases to 143 feet. Subtracting 114 from 143 indicates this is an increase of 29 feet. How do changes to the parts affect the expression? What did we change? The variable x What happened when we changed it? The Perimeter grew when we went from x=4 to x=5 1. You can use Box method, Distribution, or FOIL method to multiply the binomials. 2. Make sure you simplify your result by combining any like terms. 3. Your answer should be… 2𝑥 2 + 3𝑥 − 2 a=2 b=3 c=-2 How do changes to the parts affect the expression? What can we change? The variable, x Now let’s try substituting that x with 2 Now substitute it with -2? Bellringer 11/17/14 1. Simplify 3(12 – x2) + (3 + x) and put in standard form 2. Identify the Terms, Coefficients, and Constants. 3. Identify a, b, and c. Remember standard form of a quadratic expression is ax2 + bx + c, Are all quadratic expressions factorable? Review A factor is one of two or more numbers or expressions that when multiplied produce a given product. Example 1: 3 and 4 are factors of 12, (3*4=12) Example 2: 𝑥 + 5 2𝑥 − 7 simplified to 2𝑥 2 + 3𝑥 − 35, so 𝑥 + 5 𝑎𝑛𝑑 (2𝑥 − 7) are factors of 2𝑥 2 +3𝑥 − 35. • Sometimes it is easier to work with the factored form to determine the values that make an expression negative or equal to 0. What values of x make (x + 7)(x – 10) equal Zero? Remember: 0 times anything is 0 In order for an expression of factors to have a product of 0, only one of the factors has to be 0 What makes (x + 7) = 0? -7 What makes (x – 10) = 0? 10 What values of x make the expression (x + 2)(x – 3) positive? Determine the sign possibilities for each factor. The expression will be positive when both factors are positive or both factors are negative. Determine the values of the variable that make both factors positive. Determine the values of the variable that make both factors negative. x + 2 is positive when x > –2. x + 2 is negative when x < –2. x – 3 is positive when x > 3. x – 3 is negative when x < 3. Both factors are positive when x > –2 and x > 3, or when x > 3. Both factors are negative when x < –2 and x < 3, or when x < –2. What values of x make the expression (x + 7)(x – 10) negative? What values of x make the expression (x + 7)(x – 10) positive? A negative number is any A positive number is any number number less than 0 In order to have a negative you need 1 negative number times 1 positive number x+7 is negative when x<-7 x-10 is negative when x<10 –7 < x < 10 will result in a negative expression greater than than 0 In order to have a positive you need 2 negative numbers or 2 positive numbers x+7 is positive when x>-7 x-10 is positive when x>10 x<-7 or x >10 will result in a positive expression Multiply (x + 6)(x +3) and simplify: Multiply (x + 7)(x – 10) and simplify: (x + 7)(x – 10) (x + 6)(x +3) x2 + 7x – 10x – 70 x2 + 6x+3x + 18 x2– 3x – 70 x2 + 9x + 18 Notice: What are the sums of the constants in the original binomials? -3 Notice: What are the sums of the constants in the original binomials? -70 Notice: What is the product of the 18 constants in the original binomials? Notice: What is the product of the constants in the original binomials? 9 Bellringer 11/18/14 1. What are two factors of 18 that add up to 11? 2. What are two factors of 36 that add up to -15? 3. What is the greatest common factor of 36 and 18? Are all quadratic expressions factorable? Factoring a polynomial means expressing it as a product of other polynomials. Factoring Method #1 Factoring polynomials with a common monomial factor (using GCF). **Always look for a GCF before using any other factoring method. Steps: 1. Find the greatest common factor (GCF). 2. Divide the polynomial by the GCF. The quotient is the other factor. 3. Express the polynomial as the product of the quotient and the GCF. Example : Step 1: 2x 12 x 8 x 3 2 GCF 2x Step 2: Divide by GCF (2x 12 x 8 x) 2 x x 6 x 4 3 2 2 Step 3: Express the polynomial as the product of the quotient and the GCF. 2x(𝑥 2 − 6𝑥 + 4) Example : 6c d 12c d 3cd 3 2 2 GCF 3cd Step 1: Step 2: Divide by GCF 3 2 2 (6c d 12c d 3cd) 3cd 2 2c 4cd 1 The answer should look like this: 3 2 2 Ex: 6c d 12c d 3cd 2 3cd(2c 4cd 1) Factor these on your own looking for a GCF. 1. 6x 3x 12x 3 x 2 x 3 2 2 2. 5x 10x 35 2 x 4 5 x2 2 x 7 3. 16x y z 8x y z 12xy z 3 4 2 2 3 4 xy z 4 x y 2 xz 3 yz 2 2 2 2 3 2 Bellringer 11/19/14 Factor by finding the GCF 2 1. 15𝑥 + 2𝑥 2. 9𝑥 3 + 18𝑥 2 − 81𝑥 3. 16𝑥 3 𝑦 4 z − 8𝑥 2 𝑦 2 𝑧 3 + 12𝑥𝑦 3 𝑧 2 Are all quadratic expressions factorable? Bellringer 11/20/14 Factor 1. 12a4 - 10ab3 + 18a3 2. 70x2y3 + 56x3 + 49x 3. 10x2y3 + 4xy3 - 4x4 Are all quadratic expressions factorable? Factoring Method #2 Factoring polynomials that are a difference of squares. To factor, express each term as a square of a monomial then apply 2 2 the rule... a b (a b)(a b) 2 Ex: x 16 2 2 x 4 (x 4)(x 4) Here is another example: 1 2 x 81 49 2 1 x 92 1 x 91 x 9 7 7 7 Try these on your own: 1. x 121 2 x 11 x 11 2. 9y 169x 2 2 3 y 13 x 3 y 13 x 3. x 16 x 2 x 2 x 4 4 Be careful! 2 Hint for #3: remember your rule for raising a power to a power, and you may have to factor this one twice! Bellringer 11/20/14 Factor 1. v2 – 16 2. 9r2 – 16 3. 75n2 - 3 Are all quadratic expressions factorable? Bellringer 11/21/14 Have a seat Clear your desks of everything I mean everything! How could I have done better on the last Test? Bellringer 12/1/14 1. What is the greatest common factor of 27 and 18? 9 2. What is the greatest common factor in 4𝑥 2 − 24𝑥? 4x 3. Factor 6𝑥 2 − 30𝑥 6x(x-5) Are all quadratic expressions factorable? Factoring Technique #3 Factoring By Grouping for polynomials with 4 or more terms Factoring By Grouping 1. Group the first set of terms and last set of terms with parentheses. 2. Factor out the GCF from each group so that both sets of parentheses contain the same factors. 3. Factor out the GCF again (the GCF is the factor from step 2). Example 1: b 3 3b Step 1: Group b 3b 3 2 2 4 b 12 4b 12 Step 2: Factor out GCF from each group b2 b 3 4 b 3 Step 3: Factor out GCF 2 again b 3b 4 2 x 16 x 6 x 48 Example 2: 3 2 2 x 8 x 3 x 24 3 2 2 x 8 x 3x 24 2 x 3 2 x 8 3 x 8 2 2 x 8 x 3 2 3 2 1. 15𝑥 − 6𝑥 − 20x + 8 (3𝑥 2 − 4)(5𝑥 − 2) 3 2 2. 2𝑥 − 14𝑥 − 𝑥 + 7 (2𝑥 2 − 1)(𝑥 − 7) 3. 18𝑥 3 + 15𝑥 2 − 6𝑥 − 5 (3𝑥 2 − 1)(6𝑥 + 5) 4. 8𝑥 3 + 5𝑥 2 − 48𝑥 − 30 (𝑥 2 − 6)(8𝑥 + 5) Bellringer 12/2/14 Solve by grouping 1. 2𝑥 3 + 12𝑥 2 + 5𝑥 + 30 (2𝑥 2 + 5)(𝑥 + 6) 2. 8𝑥 3 + 5𝑥 2 − 48𝑥 − 30 (𝑥 2 − 6)(8𝑥 + 5) 3. 18𝑥 3 + 15𝑥 2 − 6𝑥 − 5 (3𝑥 2 − 1)(6𝑥 + 5) Are all quadratic expressions factorable? Factoring Method #4 Factoring a trinomial in the form: 2 ax bx c FACTOR THE X-BOX WAY We are going to factor trinomials in Standard form (ax2 + bx +c) using the X-Box method. Step 1: Write the polynomial in standard form. Step 2: Factor all common factors in the trinomial. (GCF) Step 3: Use the X method. Step 4: Write your answer. Step 5: Check your answer by distributing FACTOR THE X-BOX WAY ax2 + bx + c Sometimes m and n are easy to find. When they aren’t it is best to list out factors of ac and see which ones add up to b First and Last Coefficients Product ac=mn n m b=m+n Sum Middle EXAMPLES Factor using the x-box method. -12 6 -2 4 Factors of -12 1 -12 -1 12 2 -6 -2 6 3 -4 -3 4 x2 + 4x – 12 Which Factors add to b? 1+ -12 = -11 -1 + 12 = 11 2 + -6 = -4 -2 + 6 = 4 3 + -4 = -1 -3 + 4 = 1 Solution: x2 + 4x – 12 = (x + 6)(x - 2) EXAMPLES CONTINUED 2. x2 - 9x + 20 20 -4 -5 -9 Factors of 20 1 20 2 10 4 5 -1 -20 -2 -10 -4 -5 Which Factors add to b? 1 + 20 = 21 2 + 10 = 12 4+5=9 -1 + -20 = -21 -2 + -10 = -12 -4 + -5 = -9 Solution: x2 - 9x + 20 = (x - 4)(x - 5) 1. x2 – 6x + 5 2. x2 + 6x + 5 3. x2 – 12x + 35 4. 6x2 - 12x – 18 Bellringer 12/3/14 Factor the Trinomials: 1. x2 + 5x + 6 (x+3)(x+2) 2. x2 + 2x - 8 (x+4)(x-2) Are all quadratic expressions factorable? 3. 4x3 + 28x2 + 24x 4x(x+6)(x+1) Make sure your phone is off or on silent! If I see it, I take it up for the rest of the day! FACTOR THE X-BOX WAY Example: Factor 3x2 -13x -10 (3)(-10)= (x-15)(x+2) -30 2 -15 x -15 x x2 -15x +2 2x -30 -13 Check your work and you get x2 -13x -30 That wasn’t the original problem, … FACTOR THE X-BOX WAY Example: Factor (3)(-10)= -30 2 -15 -13 3x2 -13x -10 3x2 -13x -10 = (x-5)(3x+2) The x-box method needs help when the leading coefficient is not equal to 1. We must use the fact that the leading coefficient is 3. 15 2 3x (𝑥 − )(𝑥 + ) 3 3 Now, reduce the fractions, if possible. The coefficient of x will be the reduced denominator. +2 5 2 (𝑥 − )(𝑥 + ) 1 3 (𝑥 − 5)(3𝑥 + 2) x -5 3x2 -15x 2x -10 FACTOR THE X-BOX WAY Another Way to look at it: ax2 + bx + c = (dx+e)(fx+g) First and Last Coefficients Product dx e ac=mn fx 1st Term Factor n g Factor m Last term m n Middle b=m+n Sum EXAMPLES CONTINUED 2x2 - 5x - 7 a) b) -14 -7 2 -5 2x x +1 -7 2x2 -7x 2x -7 Solution: 2x2 - 5x – 7 = (2x - 7)(x + 1) EXAMPLES CONTINUED 15x2 + 7x - 2 a) b) -30 10 -3 7 3x +2 5x 15x2 10x -1 -3x -2 Solution: 15x2 + 7x – 2 = (3x + 2)(5x - 1) 1. 2x2 – 19x + 24 (x-8)(2x-3) 2. 2x² + 13x + 6 (x + 6)(2x + 1) 3. 6x2 + x -15 (3x + 5)(2x – 3) 4. 6x2 + 13x -5 (2x + 5)(3x – 1) Bellringer 12/4/14 Factor the Trinomials: 1. x2 - 9x + 18 (x - 6)(x - 3) 2. 2x2 - 9x - 18 Are all quadratic expressions factorable? (x - 6)(2x + 3) 3. 9x2 - 4 (3x - 2)(3x + 2)] Make sure your phone is off or on silent! If I see it, I take it up for the rest of the day! Factoring Technique #5 Factoring a perfect square trinomial in the form: a 2ab b (a b) 2 2 a 2ab b (a b) 2 2 2 2 Perfect Square Trinomials can be factored just like other trinomials, but if you recognize the perfect squares pattern, follow the formula! a 2ab b (a b) 2 2 a 2ab b (a b) 2 2 2 2 2 Ex: x 8x 16 x Does the middle term fit the pattern, 2ab? 4 2 2 a b 2 x 4 8x Yes, the factors are (a + b)2 : x 8x 16 x 4 2 2 2 Ex: 4x 12x 9 2x Does the middle term fit the pattern, 2ab? 3 2 2 a b 2 2x 3 12x Yes, the factors are (a - b)2 : 4x 12x 9 2x 3 2 2 1. x2 – 6x + 9 (x - 3)2 2. 9x2 + 12x + 4 (3x + 2)2 3. 16x2 – 56x + 49 (4x - 7)2 4. 4x2 + 20x + 25 Bellringer 12/5/14 Factor the following: 1. - 40x2y - 8y - 8y (5x2 + 1) Are all quadratic expressions factorable? 2. 9x2 - 12x + 4 (3x - 2)2 3. 25n2 - 9 (5n + 3)(5n - 3) 4. 12p3 + 8p2 + 21p + 14 (4p2 + 7)(3p + 2) Make sure your phone is off or on silent! If I see it, I take it up for the rest of the day! First group that has every question answered correctly on everyone’s paper with work shown wins 5 bonus Points on the test Tuesday! 2nd group gets 4 points, 3rd gets 3, 4th gets 2, and 5th gets 1 Bellringer 12/8/14 Take out your Review sheet that you should have completed this weekend. Make sure your phone is off or on silent. If I see it, I take it up for the rest of the day! Are all quadratic expressions factorable? Bellringer 12/9/14 1. What factoring method do you use when you have 4 terms? 2. What factoring method should we try to use if there are 2 terms? 3. What factoring methods can we use when we have 3 terms When you are done clear your desks except for a pencil, calculator, graphic organizer, and 1 sheet of paper to show your work if you would like. Make sure your phone is off or on silent! Factoring a trinomial using guess and check: 2 ax bx c 1. Write two sets of parenthesis, ( )( ). These will be the factors of the trinomial. 2. Product of first terms of both binomials 2 must equal first term of the trinomial. ( ax ) Next Factoring a trinomial using guess and check : 2 ax bx c 3. The product of last terms of both binomials must equal last term of the trinomial (c). 4. Think of the FOIL method of multiplying binomials, the sum of the outer and the inner products must equal the middle term (bx). x 6x 8 2 Example : x x -2 x x Factors of +8: -4 x x x O + I = bx ? 1&8 1x + 8x = 9x 2&4 2x + 4x = 6x -1 & -8 -1x - 8x = -9x -2 & -4 -2x - 4x = -6x 2 x 6x 8 (x 2)(x 4) 2 Check your answer by using FOIL F 2 O I L (x 2)(x 4) x 4x 2x 8 2 x 6x 8 Lets do another example: 2 6x 12x 18 Don’t Forget Method #1. Always check for GCF before you do anything else. 2 6(x 2x 3) Find a GCF 6(x 3)(x 1) Factor trinomial When a>1 and c<1, there may be more combinations to try! Example : 6 x 13x 5 2 Step 1: 2 Find the factors of 6x : 3x 2x 6x x Example : 6 x 13 x 5 2 Step 2: 5 -1 -5 1 -1 5 1 -5 Find the factors of -5: Order can make a difference! Example : 6 x 13 x 5 2 Step 3: Place the factors inside the parenthesis until O + I = bx. Try: F 2 6x 1x 5 O I L 6x 30x x 5 O + I = 30 x - x = 29x This doesn’t work!! Example : 6 x 13 x 5 2 Switch the order of the second terms and try again. 6x 5x 1 F 2 O I L 6x 6x 5x 5 O + I = -6x + 5x = -x This doesn’t work!! Try another combination: Switch to 3x and 2x (3x 1)(2x 5) F 2 O I L 6x 15x 2x 5 O+I = 15x - 2x = 13x IT WORKS!! 6x 13x 5 (3x 1)(2x 5) 2 Bellringer 12/10/14 1. Factor: 12x3-4x2+18x-6 2. Factor: 9x2-81 3. Factor: 2x2-6x+5 When you are done clear your desks of everything except your bellringer sheet! Make sure your phone is off or on silent! Bellringer 12/11/14 Solve for x 1. x+8=0 2. 3x - 2 = 0 3. 5x + 6 = - 5 - 3x Make sure your phone is off or on silent! How do the factors of a quadratic functions yield the zeros for that function? Solving Quadratic Equations by Factoring. A quadratic equation is an equation that can be written in the form ax2 + bx + c = 0, where a, b, and c are real numbers, with a ≠ 0. The form ax2 + bx + c = 0 is the standard form of a quadratic equation. For example, x 2 5x 6 0, 2 x2 5x 3, and x 2 4 are all quadratic equations, but only x2 + 5x +6 = 0 is in standard form. Until now, we have factored expressions, including many quadratic expressions. In this section we see how we can use factored quadratic expressions to solve quadratic equations. Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley We use the zero-factor property to solve a quadratic equation by factoring. If a and b are real numbers and if ab = 0, then a = 0 or b = 0. That is, if the product of two numbers is 0, then at least one of the numbers must be 0. One number must, but both may be 0. Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 6.5 - 91 Using the Zero-Factor Property Solve. Solution: 2x 35x 7 0 3 7 , 2 5 2 x 3 0 or 5x 7 0 2x 3 3 0 3 5x 7 7 0 7 2 x 3 5 x 7 2 2 5 5 3 7 x x 2 5 x 2 x 4 0 0, 2 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley x0 or 2x 4 0 2x 4 4 0 4 2 x 4 2 2 x 2 Slide 6.5 - 92 Bellringer 12/12/14 Solve for x 1. (x + 5)(x - 4) = 0 2. (7x - 1)(x+9)= 0 3. x2 + 3x + 2= 0 Make sure your phone is off or on silent! How do the factors of a quadratic functions yield the zeros for that function? Solving Quadratic Equations Solve. x2 2 x 8 x2 2x 8 8 8 x2 2x 8 0 x 4 x 2 0 Solution: x40 x2 0 or x44 04 x 22 02 x2 x 4 4, 2 x5 0 x 2 x 30 or x6 0 x 55 05 x2 x 30 x 30 x 30 x 6 6 0 6 x6 x 5 x 2 x 30 0 x 6 x 5 0 5,6 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 6.5 - 94 In summary, follow these steps to solve quadratic equations by factoring. Step 1: Write the equation in standard form— that is, with all terms on one side of the equals sign in descending power of the variable and 0 on the other side. Step 2: Factor completely. Step 3: Use the zero-factor property to set each factor with variable equal to 0, and solve the resulting equations. Step 4: Check each solution in the original equation. Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 6.5 - 95 Solving a Quadratic Equation with a Common Factor 2 Solve 3m − 9m = 30. m2 0 m 2 2 0 2 Solution: m 2 3m2 9m 30 30 30 2 3m 9m 30 0 m5 0 2 3 m 3m 10 0 m55 05 3 m 2 m 5 0 m5 2,5 A common error is to include the common factor 3 as a solution. Only factors containing variables lead to solutions. Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 6.5 - 96 Solving Quadratic Equations Solve. 49 x 2 9 0 7 x 3 3 0 3 Solution: 7 x 3 7 x 3 0 3 3 , 7 7 x 2 3x x 2 3x 3 x 3 x x x 3 0 7x 3 3 0 3 7x 3 7 7 3 x 7 7 x 3 7 7 x0 x 33 0 3 0,3 x 3 7 x 3 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 6.5 - 97 Solving Quadratic Equations (cont’d) x20 x22 02 x 2 Solve. x 4x 7 2 Solution: 4x 7x 2 2 2 2 4x 7x 2 0 x 2 4x 1 0 2 1 2, 4 4 x 1 0 4 x 1 1 0 1 4x 1 4 4 1 x 4 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 6.5 - 98 Solving Equations with More than Two Variable Factors 2 x3 50 x 0 Solution: 2 x x 2 25 0 Solve. 2 x x 5 x 5 0 2x 0 2x 0 2 2 x0 x 55 05 x 5 x 55 05 x 5 0, 5,5 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 6.5 - 99 Solving Equations with More than Two Variable Factors (cont’d) Solve. 2 x 1 2 x 2 7 x 15 0 Solution: 2 x 1 2 x 3 x 5 0 2 x 1 1 0 1 2 x 1 2 2 1 x 2 2x 3 3 0 3 2x 3 2 2 3 x 2 x 55 05 x 5 1 3 5, , 2 2 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 6.5 - 100 Solving an Equation Requiring Multiplication before Factoring x 1 2 x 1 x 1 Solve. 2 2 x 2 3x 1 x 2 2 x 1 2 2 2 2 2 x 3x 1 x 2 x 1 x 2 x 1 x 2 x 1 x2 5x 0 x x 5 0 Solution: x0 x 55 05 x 5 0,5 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 6.5 - 101 Bellringer 12/15/14 Solve for x 1. 3x + 10 = 2x - 5 2. x2 + 5x – 50 = 0 3. 2x2 – 3 = - 5x None Bellringer 12/16/14 Solve for x 1. x2 + 2x + 10 = 0 None 2. 3x2 + 9x – 12 = 0 3. 2x2 – 5x +3= 0 SOLVING QUADRATIC EQUATIONS BY THE QUADRATIC FORMULA THE QUADRATIC FORMULA 1. When you solve using completing the square on the general formula ax 2 bx c 0 you get: b b2 4ac x 2a 2. This is the quadratic formula! 3. Just identify a, b, and c then substitute into the formula. WHY USE THE QUADRATIC FORMULA? The quadratic formula allows you to solve ANY quadratic equation, even if you cannot factor it. An important piece of the quadratic formula is what’s under the radical: b2 – 4ac This piece is called the discriminant. The discriminant tells you the number and types of answers (roots) you will get. The discriminant can be +, –, or 0 which actually tells you a lot! Since the discriminant is under a radical, think about what it means if you have a positive or negative number or 0 under the radical. WHAT THE DISCRIMINANT TELLS YOU! Value of the Discriminant Nature of the Solutions Negative 2 imaginary solutions Zero 1 Real Solution Positive – perfect square 2 Reals- Rational Positive – non-perfect square 2 Reals- Irrational EXAMPLE #1 Find the value of the discriminant and describe the nature of the roots (real,imaginary, rational, irrational) of each quadratic equation. Then solve the equation using the quadratic formula) 1. 2 x 2 7 x 11 0 Discriminant = a=2, b=7, c=-11 b 2 4ac (7) 2 4(2)( 11) 49 88 Discriminant = 137 Value of discriminant=137 Positive-NON perfect square Nature of the Roots – 2 Reals - Irrational EXAMPLE #1- CONTINUED Solve using the Quadratic Formula a 2, b 7, c 11 2 x 2 7 x 11 0 b b 2 4ac 2a 7 7 2 4(2)(11) 2(2) 7 137 4 2 Reals - Irrational Try the following examples. Do your work on your paper and then check your answers. 1. x 2 x 63 0 2 2. x 8 x 84 0 2 3. x 5 x 24 0 2 4. x 7 x 13 0 2 5. 3 x 2 5 x 6 0 1. 9, 7 2.(6, 14) 3. 3,8 7 i 3 4. 2 5 i 47 5. 6 Bellringer 12/17/14 Solve for x by Completing the Square 1. x2 + 10x + 16 = 0 Solve for x by using the Quadratic Formula 2. 2x2 + x - 1 = 0 None