lsolve. 25(x + 3)2-2 = 0

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1 II nrm!: lsolve. 25(x + 3)2-2 = 0 ISolve. 4(x - 7) 2-5 = 0 Isolate the squared term. Move everything but the term being squared to the opposite side of the equal sign. Use opposite operations. Isolate the squared term. 25(x + 3) 2-2 = 0 25(x + 3) = (x + 3)2 _ (x + 3) = 25 Take the square root of both sides and simplify. The square root of a number could be either positive or negative. \/4 cou]d be either 2 or - 2 since 22 = 4 and (-2) 2 = 4. Take the square root of both sides and simplify. '.!4 = :t2 J(x + 3)2 = g x+3 = :t1 Solve the equation using opposite operations. Solve the equation using opposite operations. x+3 = :t1.j2 x = -3 :t 5 /2.f2 x = or , Renaissance Learning, Inc.

2 ISolve. r - lox + 24 = 0 I Solve. r + 12x + 27 = 0 Review solving quadratic equations. If x(x + 2) = 0, then either x or x + 2 must equal zero. The only way to get a product of 0 is for one or more of the factors to be equal to O. To solve a quadratic equation, always set it equal to zero and then factor it For this problem, the equation is already set equal to zero. Review solving quadratic equations. To solve a quadratic equation, make sure the equation is set equal to _~_ When a quadratic equation is factored, one or more of the factors must be ~ Factor. This expression can be factored into the form (x + a) (x + b) where ab = 24 (the constant term) and a + b = -10 (the coefficient on the middle term). Since the coefficient is negative and the constant is positive, a and b must both be negative. The product of two negative numbers is positive. The sum of two negative numbers is negative. Options for a and bare: -1 and and -8-2 and and -6-4 and -6 are the only pair with a sum of -10. r - lox + 24 = 0 (x - 4) (x - 6) = 0 Factor. This expression can be factored into the form (x + a)(x + b). ab= a+b= --- a and b must both be r + 12x + 27 = ( ) ( ) = 0 Solve each case. x-4=0 or x-6=0 x-4+4=0+4 x-6+6=0+6 x=4 x=6 Solve each case. x = 4 or 6 x= or Renaissance Learning, Inc.

3 II Objective-Factor to S4)lve quadratic equation - x2 X 3 -~ ISolve = 2 Set the equation equal to O. x 2 X x 2 X =0 Simplify the equation. Eliminate the fractions from the equation. Multiply by the least common denominator (4). 2 x 4( x = 0) 2r+x-6=0 Factor. This expression can be factored into the form (ax + b) (ex + tf) where bd = 6 (the constant term) and ae = 2 (the coefficient on the first term). Options for band d are: Options of a and e are: -land6 land-6 land2 -land-2-2 and 3 2 and -3 Use trial and error to find out which numbers work. (2x - 2) (x + 3) = 2r + 6x - 2x - 6 =2r+4x-6 (2x - 3) (x + 2) = 2r + 4x - 3x - 6 =2r+x-6 2r+x-6=0 (2x - 3) (x + 2) = 0 Solve each case. 2x-3=0 or x + 2 = 0 2x - 3+3=0+3 x+2-2=0-2 2x 3 2 = 2 x = -2 x = ~ x = 3 2 or , Renaissance Lea rning, Inc. 2 I X Solve. 2 x + '3 = '6 1 Set the equation equal to O. Simplify the equation. Factor. This expression can be factored into the form (ax + b)(ex + tf). bd= ae = Options for band d: Options of a and e: Use trial and error to find out which numbers work. Solve each case. x = or

4 II,-Find the missing terms. :C - 6x + = ( ) 2 lfudthe missing terms. x x + = (?1 Review terms. The expression "completing the square" involves altering an equation to make it a perfect square trinomial in the form (a + by/.. Remember the formula. (a + b)2 = a 2 + 2ab + b 2 In order for the method of completing the square to work, the coefficient on the x-squared term must be 1. That is already the case with this problem. If the coefficient were 2, you would start by dividing all terms by 2. Review terms. If the coefficient on the x-squared term is not 1, what would be the first step to completing the square? Find half of the coefficient on the middle term. Find half of the coefficient on the middle term. The middle term is -6x. Its coefficient is ;- 2 = -3 The middle term is Its coefficient is Half the coefficient is --- Square the number found in Step 2 and use it as the constant term. (-3f = 9 X 2-6x+ 9 Square the number found in Step 2 and use it as the constant term. :C + 14x Write the squared expression. Use x as a. Use the number found in Step 2 as b. The operation sign is the sign on the middle term. For this equation, the operation is subtraction. x ~ - 6x + 9 = (x - 3f ~ Write the squared expression. Use for a. Use as b. The operation is ( )2 Answer: x 2-6x + 9 (x - 3)2 Answer: x x + = ( )2 2002, Renaissance Learning. Inc.

5 I Solve by completing the square. 9x = S,r - 1 I Sol;e by completing the square. -6x- = 3x 2-2 Put all x terms on one side of the equal sign and constant(s) on the other side. 9x = Sx 2-1 9x - 9x = SX l - 9x - 1 o = Sx 2-9x - 1 o+ 1 = Sx 2-9x = SX l - 9x Put all x terms on one side of the equal sign and constant(s) on the other side. Divide the equation by the coefficient on the x-squared term.! (1 = si - 9x) Divide the equation by the coefficient on the x-squared term x :5 =X - 5 Square half of the coefficient on the x term. Square half of the coefficient on the x term. - ~ is the coefficient on the x term. 9 )2 (9 1)2 ( 9)2 81 ( - : = -:5' 2 = - 10 I = 100 is the coefficient on the x term. ( -7-2) 2 = Add the number found in Step 3 to both sides of the equation x :5 =X - 5 Add the number found in Step 3 to both sides of the equation x 81 : = x x = x x =x

6 I Solve by co~pleting the square. 9x = 5r - 1 lsolve by completing the square. -6x = 3x 2-2 Factor the trinomial. Use the formula if you need a reminder. a -x - b- -..JiOO ill - JL 10 Factor the trinomial. a= b= (a + b)2 = a + 2ab + 2 b 2 9 )2 2 9x 81 (x - 10 = x x = x ( 9 )2 100 = x - 10 Step 6: Solve the equation using the square root rule. 101 t 9 )2 100 = \X - TO JWo - (x 10) +.JIill x _ JL lill- J _J! 'J +YTIIT x- lo lo Step 6: Solve the equation using the square root rule. If the solution has a fraction with a square root in the denominator, remember to rationalize the denominator. 9 ±-V'IOi x v'ioi 9 --v'ioi Answer: x 10 or 10 Answer: x = or , Renaissance Learning. Inc.

7 II ISolve. 8x = 3r + 3 Review the quadratic formula. x -b:t: Jb 2-4ac 2a You can find the value of x in any quadratic equation using this formula as long as the equation is in the form ar + bx + c = O. Set the equation equal to O. 8x=3r+3 8x-8x= 3r-8x+3 O=3,r-8x+3 Identify a, b, and c. a is the coefficient on the x-squared term. a=3 b is the coefficient on the x term. b = -8 c is the constant. c = 3 \ Solve. 8x = 5r + 3 \ Review the quadratic formula. Write the quadratic formula. Set the equation equal to o. Identify a, b, and c. a= b= c= Substitute the values for a, b, and c into the formula and simplify. x x -b:t: Jb 2-4ac 2a - (-8) :t: j '---( ---:8)2-.,---(--:4)--:-(3-:-)("""' 3 ) 2(3) Substitute the values for a, b, and c into the formula and simplify. _ 8:t: )64-36 _ 8 ± J28 _ 8:t: 2/ Since all the terms have a factor of 2, you can cancel it out. 2(4 ± vi) 4+ft 2(3) 3 Answer: x 4+ft 4-ft -3- or -3 Answer: x = or , Renaissance Learning, Inc.

8 II For what values of b does y = 27x2 + bx + 3 have two x-intercepts? For what values of c does y = - one x-intercept? 2x 2 + 4x + c have exactly Review intercepts. A graph intersects the x-axis when y = o. Replacing y with 0 and solving for x will tell you the x-intercepts. A graph intersects the y-axis when x = o. Replacing x with 0 and solving for y will tell you the y-intercepts. Review intercepts. What is the value of y for each x-intercept? What is the value of x for each y-intercept? Find the discriminate. Ji2 The quadratic formula is x = -b::!::: ~ - 4ac.. The discriminate is the part of the quadratic formula under the square root sign: b 2-4ac. The discriminate can tell you how many x-intercepts there are. In the quadratic equation above, a = 27, b is unknown, and c = b - 4ac = b - 4(27)(3) = b Find the discriminate. b 2-4ac = Evaluate the discriminate. discriminate> 0 discriminate = 0 discriminate < 0 there are two x-intercepts there is one x-intercept there are no x-intercepts The question asks what the values of b must be for there to be two x-intercepts, so we want a discriminate that is greater than o. Evaluate the discriminate. Describe the discriminate when there is exactly one x-intercept. Write and solve an inequality or equation. Write an inequality that makes the discriminate greater than o. b 2-324> 0 b > 324 Iil > 54 Write and solve an inequality or equation. b > 18 or b < Renaissance learning. Inc.

9 [ Gr;ph. y = x 2-4x - 2 [ Graph. y = r - 2x + 3 Review quadratic equations. The graphs of all quadratic equations are parabolas. Parabolas are curved shapes that look like bowls or arches, depending on which direction they open. You can graph a quadratic equation by evaluating a, b, and c. For this equation, a = 1, b = -4, and C = -2. Review quadratic equations. a= b= c= Plot the vertex. The vertex is the point at the bottom of the bowl or the top of the arch of the parabola. Use the formula to find the vertex. Plot the vertex. Vertex = (2:' f( 2:)) y. - b a = - 2(1) = 2 = 2 Find the y-value of the equation when x = 2.. : x y = x 2-4x - 2 = 22-4(2) - 2 = = - 6 vertex The vertex is (2, - 6). The vertex is ~ Use the discriminate to determine the number of x-intercepts. b'2-4ac = (_4)2-4(1)(- 2) > 0: two x-intercepts = 16 - (-8) = 0: one x-intercept = = 24 < 0: no x-intercepts Use the discriminate to determine the number of x-intercepts. This parabola has two x-intercepts. This parabola has x-intercepts.

10 2 IGraph. y = x - 4x J Graph. y = r - 2x + 3 Determine the x- and y-intercepts. Solve for y = 0 using the quadratic formula to find the x-intercepts. Determine the x- and y-intercepts. x - (-4) ±y(-4r - (4)(1)(- 2) (2)1 4±J16 - (-8) = 4±) )24 4-)24 The x-mtercepts are at 2 and ~. Use a calculator to find decimal approximations of these values. 4 + f )24 :::: -0.4 Plot points at (4.4,0) and (-0.4,0). Solve for x = 0 to find the y-intercepts. y = (0) 2-4(0) - 2 = = -2 The y-intercept is at - 2. Plot a point at (0, - 2). Connect the points to draw the graph. All parabolas are symmetric. Make the right side coming up froid the vertex sym-.metrical to the left side. y. Connect the points to draw the graph. All parabolas are symmetric. Make the right side coming up from the vertex symmetrical to the left side. y.. 1. x t x 2002, Renaissance Learning, Inc.

11 II For the parabola, state the direction of the opening, the coordinates of the vertex, the equation of the axis of symmetry, and the maximum or minimum value. y = -0.5(x - 3) For the parabola, state the direction of the opening, the coordinates of the vertex, the equation of the axis of symmetry, and the maximum or minimum value. y = 0.8(x + 4) Put the equation in standard form. Put the equation in stand_ard form. Quadratic equations in standard form look like this: y = ax 2 + bx + C Y = -O.5(X - 3) = -0.5(r - 6x + 9) + 5 = -0.5r + 3x = -O.5r + 3x For this equation, a = -0.5, b = 3, and c = 0.5. b= a= c= Determine if the parabola opens upward or downward. When a > 0, the parabola opens upward. When a < 0, the parabola opens downward. Since a = -0.5, this parabola opens downward. Determine if the parabola opens upward or downward. This parabola opens Find the coordinates of the vertex. The vertex is the point at the bottom of the bowl or the top of the arch. Use the formula to find the vertex. Vertex = (2: ' f( 2:)) Find the coordinates of the vertex. -b a = 2(-0.5) = -1 = Y = -O.5i + 3x = -0.5(3 2 ) + 3(3) = -0.5(9) = =5 The vertex is (3, 5). The vertex is

12 For the parabola, state the direction of the opening, the coordinates of the vertex, the equation of the axis of symmetry, and the maximum or minimum value. 2 y = -O.5(X - 3) + 5 For the parabola, state the direction of the opening, the coordinates of the vertex, the equation of the axis of symmetry, and the maximum or minimum value. y = 0.8 (x + 4r 'J + 5 Find the equation of the axis of symmetry. The axis of symmetry is a line of symmetry for the parabola. The equation for the axis of symmetry Find the equation of the axis of symmetry. The equation for the axis of symmetry. -b IS x = 2a' IS In Step 3, you found the value of 2: to be 3. So, the equation for the axis of symmetry is x = 3. Find the minimum or maximum value. When a parabola opens upward, y is at its minimum value at the vertex. When a parabola opens downward, y is at its maximum value at the vertex. This parabola opens downward. Its vertex is at (3, 5). Its maximum value is at y :;;;:: 5. Find the minimum or maximum value. Because this parabola opens it has a value at , Renaissance Learning, Inc.

13 Use the graph of y = X2, shown on the left, to find the equation for the graph on the right. y. y. " I" x ~ x n Use the graph of y = x 2, shown on the left, to find the equation for the graph on the right. y. u >'1" x. x Reflect the direction change in the equation.. The parabola switched from opening upward to opening downward. Remember that the direction is determined by a, the coefficient on i. When a is negative, the parabola opens downward. If the only change in the graph were that it opened downward, the equation would be y = -x 2 Reflect the direction change in the equation. Did the direction of the parabola's opening change? Should a be positive or negative? Reflect the change in the position of the vertex in the equation. The vertex of the parabola has shifted 3 units on the x-axis and - 2 units on the y-axis. To shift the position of the vertex on the x-axis, add or subtract from x before it is squared. For example, y = (x + 1)2causes the vertex to shift 1 unit in the negative direction on the x-axis. This graph shifted 3 units in the positive direction on the x-axis, so subtract 3 from x. y = (x - 3)2 Reflect the change in the position of the vertex in the equation. This graph shifted unites) in the direction on the x-axis, so x. y=

14 Use the graph of y = X2, shown on the left, to find the equation for the graph on the right. y. y. Use the graph of y = x 2, shown on the left, to find the equation.. for the graph on the right. y. y. '1'/ II x ox n ' I <' ox ~ ~ox (Continued) To shift the position of the vertex on the y-axis, add or subtract a value to the squared value. For example, y = r + 1 causes the vertex to shift 1 unit in the positive direction on the y-axis. This graph shifted 2 units in the negative direction on the y-axis, so subtract 2 from the squared term. 2 y = -(x - 3) - 2 (Continued) This graph shifted unites) in the direction on the y-axis, so the squared term. y = Reflect the change in the width of the opening of the parabola. Imagine what the parabola would look like if only the opening had narrowed. The first points above the vertex would have been (1, 2) and (-1, 2). In the original graph, the first two points above the vertex are (1, 1) and (-1, 1). The y values are doubled in the second graph. To double the value of y, multiply the squared term by 2. y = zr Reflect the change in the width of the opening of the parabola. If only the opening had narrowed, the first points above the vertex would have been (1, ) and (-1, ). The y values are To reflect this change, multiply the squared term by Answer: The equation of the graph on the right is y = -2(x - 3) 2-2. Answer: The equation of the graph on the right is 2002, Renaissance Learning, Inc.

15 II One leg of a right triangle is 31 inches longer than the other leg. The hypotenuse is 41 inches long. How long is the shorter of the legs? One leg of a right triangle is 9 inches longer than the other leg. The hypotenuse is 45 inches long. How long is the shorter of the legs? Review the Pythagorean theorem. 2 a + b 2 = c, where a and b represent the ~ c lengths of two legs of a right triangle and a ~ c represents the length of the hypotenuse. b Review the Pythagorean theorem. Write the Pythagorean theorem. Describe the lengths of the three sides of the triangle. Let s represent the length of the shorter leg. a = shorter leg = s The longer leg is 31 inches longer. b == longer leg = s + 31 The hypotenuse is 41 inches long. c = hypotenuse ::;;: 41 Describe the lengths of the three sides of the triangle. a = shorter leg = s b = longer leg = c = hypotenuse = _ Substitute what you know into the formula and simplify the equation. Substitute what you know into the formula and simplify the equation. a 2 + b 2 = c 2 S2 + (s + 31)2 = 412 S2 + S2 + 62s = S2 + 62s = 0 Solve the equation using the quadratic formula. For the quadratic formula, a = 2, b = 62, and c = _ -62:t j s - (2)2 4(2)(-720) Solve the equation using the quadratic formula. a= b= c = -62:tJ (- 5760) _ -62:tj% s = 9 or -40 s = or Answer the question. Length can only be positive. Therefore, the length of the shorter leg is 9 inches. Answer the question. The length of the shorter leg is 2002, Ren aissance Lea rn ing. Inc.

16 II The product of two consecutive positive integers is 482 more than the next integer. What is the largest of the three integers? The product of two consecutive positive integers is 23 more than the next integer. What is the largest of the three integers? Review terms. :Consecutive means numbers next to each other when counting. 4, 5, and 6 are three consecutive positive integers. This problem states that the product of two consecutive integers is larger than the next integer. The next integer is one more than the larger of the two multiplied integers, or two more than the smaller of the two multiplied integers. For example, if the two integers multiplied were 4 and 5, the next integer would be 6 and 6 is the greatest of the three integers. Review terms. Give an example of three consecutive, positive integers. Describe the numbers. Let s represent the least of the three consecutive integers. The second consecutive integer is one more than the first. It can be described by s + l. The third consecutive integer is two more than the first. It can be described as s + 2. Describe the numbers. Let s represent the least of the three consecutive integers. The second consecutive integer can be described as The third consecutive integer can be described as

17 The product of two consecutive positive integers is 482 more than the next integer. What is the largest of the three integers? The product of two consecutive positive integers is 23.more than the next integer. What is the largest of the three integers? Write and simplify an equation. You know that the product of the two consecutive integers is 482 more than the third integer. Write and simplify an equation. The product of the two integers = The product of the two integers = s(s + 1) 482 more than the third integer = (s + 2) s(s + 1) = (s + 2) s 2 + S = s + 2 S = s - s + 2 S2 = 484 S2 + S - 23 more than the third integer = The entire equation is: Solve the equation. Solve the equation. s2 = 484 JS2 = )484 s = ± 22 Answer the question. Answer the question. The problem described positive integers, so you can throw out The greatest of the three integers is s + 2 = = 24. The greatest of the three integers is Renaissance Learning, Inc.

18 The perimeter of a rectangular sheet of paper is 52 inches and its area is 165 square inches. What is the length of the longer side of the sheet of paper? The length of a rectangle is 7 centimeters less than two times its width. If the area is 30 square centimeters, find the length and width. Review formulas. P = 2l + 2w A = lw Review formulas. What formula will you need to use for this problem? Describe the length and width. You know that the area is 165 in 2 You can describe the width in terms of the length using this equation. 165 = lw 165 _ lw I - I Describe the length and width. Let w represent the width. Write the phrase "7 centimeters less than two times its width" as an algebraic expression. This expression will describe the length in terms of the width w I Write and simplify an equation. Substitute the value for w in terms of l into the perimeter formula. P = 2l + 2w = 52 Write and simplify an equation. 2l + 2e~5) = 52 2l = 52 I l(2l + 3~0 = 52) To get lout of the = 52l denominator, multiply the entire equation by l l = 0 Please turn the card over for the rest of the probl_em.

19 The perimeter of a rectangular sheet of paper is 52 inches and its area is 165 square inches. What is the length of the longer side of the sheet of paper? The ~ength of a rectangle is 7 centimeters less than two times its width. If the area is 30 square centimeters, find the length and width. Solve the equation. Solve the equation. a = 2, b = -52, and c = 330 a= b= c= l = - (- 52) ::t: Jc=52)2-4 (2)(330) 2(2) = 52::t: / = 52::t: / ::t: 8-4 l :;;; 11 or 15 w= or Answer the question. Answer the question. Since the length and the width could easily be swapped in the two formulas, the two values for l are the length and the width. The length of the longer side of the sheet of paper is 15 in. The width is The length is 2002, Renaissance Learning, Inc.

20 Suppose you throw a ball into the air with an initial velocity of 32 feet per second. The equation that gives its height after t seconds is h = -16t t + 5. Find the maximum height the ball will reach and the time it will take for the ball to reach that height. Carlin drove 756 miles round trip to visit his friend in Bellows Falls. He averaged 12 miles per hour faster on the return trip because he took a highway route. The round trip took 16 hours to drive. At what average speed did Carlin drive on the way to and from his friend's home? Review parabolas. The shape of the path the ball will make is a parabola. It will go up into the air, slow down as it reaches its maximum height, and then fall back to the ground picking up speed as it drops. The maximum height it reaches is the value of y at the vertex of the parabola. y in this equation is represented by h. Vertex Maximum height Review distance problems. The round trip was 756 miles, so each leg of the trip was half that, or 378 miles. Let r represent the rate on Carlin's ride to his friend's house. Since Carlin averaged 12 miles per hour faster on the return trip, the return rate can be represented by Let t represent the time it took Carlin to get to his friend's house. Since the entire trip took 16 hours, the return time can be represented by To Distance Rate Time I From - Find the vertex. Use the formula to find the vertex. Vertex = (2:'/ ( 2: )) For the given equation, a = -16 and b = 32. -b = 2 ( - 16) = - 32 = 1 Write an equation and simplify. Distance = rate. time Distance (to): 378 = Distance (from): 378 = Combine the equations into one and simplify. rt =

21 Suppose you throw a ball into the air with an initial velocity of 32 feet per second. The equation that gives its height after t seconds is h = -16t t + 5. Find the maximum height the ball will reach and the time it will take for the ball to reach that height. Carlin drove 756 miles round trip to visit his friend in Bellows Falls. He averaged 12 miles per hour faster on the return trip because he took a highway route. The round trip took 16 hours to drive. At what average speed did Carlin drive on the way to and from his friend's home? (Continued) Find h when t is 1. h = - 16t t + 5 = -16(1)2 + 32(1) + 5 = ~ = 21 The vertex of the parabola is (1,21). (Continued) To solve an equation, it must have only one variable. Find the value of t with respect to r in the equation 378 = rt. You know the value of rt is 378. Substitute 378 for rt and for t. Multiply by r to get r out of the denominator. Answer the question. Solve the equation and answer the question. The vertex of this parabola is (1,21) where 1 is a value of t, time, and 21 is a value of h, height. The ball reaches its maximum height of 21 feet in one second. r= or Carlin drove an average speed of per hour on the way to his friend's house and miles miles per hour on the return trip. 2002, Renaissance Learning, Inc.

22 When Alex, Karen, and Joe work together to paint a house, they can finish it in 2 days. Working alone, Alex can paint a house in 5 days, while Karen can do it in 1 day less than Joe. How long would it take Joe to paint a house by himself? (Round your answer to the nearest tenth.) When Nick, Pauline, and BJ work together to do the yard work for an apartment complex, they can finish the job in 8 hours. Working alone, Nick can complete the job in 20 hours, while Pauline can do it in 3 hours less time than B1. How long would it take BJ to do all the yard work by herself? (Round your answer to the nearest tenth.) Review work problems. The equation used to describe shared work is always set equal to 1 to represent one job completed. A fraction is used to represent each person's contribution to work done. Review work problems. When you write an equation for a work problem, it is set equal to 1 because Find fractions that represent each person's share of the work. Alex can paint a house in 5 days. In one day, he would have! of the job done. When they work together, they complete a house in 2 days. In two days, Alex would have ~ of the job done. Find fractions that represent each person's share of the work. Nick can do the yard work in hours. In one hour he would have of the work done. In the We don't know anything about how fast Joe works. Let's use j to describe the number of days it would take for him to paint a house. In two days, he'd have J of the job done. Karen can complete the job in 1 day less than it takes Joe. If it takes Joe j days, it will take Karen j - 1 days. In two days, Karen would have j -= 1 of the job done. time it takes all three to complete the job, hours, he would have of the work done. Choose a variable to represent the number of hours it would take BJ to complete the yard work by herself. Write a variable expression for the part of the job completed when they work together. BJ= Pauline =

23 When Alex, Karen, and Joe work together to paint a house, they can finish it in 2 days. Working alone, Alex can paint a house in 5 days, while Karen can do it in I day less than Joe. How long would it take Joe to paint a house by himselt? (Round your answer to the nearest tenth.) When Nick, Pauline, and BJ work together to do the yard work for an apartment complex, they can finish the job in 8 hours. Working alone, Nick can complete the job in 20 hours, while Pauline can do it in 3 hours less time than BJ. How long would it take BJ to do all the yard work by herselt? (Round your answer to the nearest tenth.) Write an equation and simplify. Alex's + Joe's + Karen's I house work work work painted 222 "5 + J + j-l I Multiply the entire equation by 5j(j - 1) to get these terms out of the denominator. 5j (j - 1) [~ + J + j ~ 1] = I Write an equation and simplify. Nick's + BJ's + Pauline's ljob work work work done + + Multiply the entire equation by to get these terms out of the denominator. I 2j(j - 1) + 1O(j - 1) + loj = 5j(j - 1) 2j2-2j + loj loj = 5F - 5j 2F + 18j - 10 = 5F - 5j - 3j j - 10 = 0 Solve the equation. a = - 3, b = 23, and c = -10 J = -23:!: J (-3)(-10) 2(-3) Solve the equation. a= b= c= ~ 23:!:J :!:j :!: J = 0.5 or 7.2 Answer the question. It doesn't make sense that Joe could paint the house in less time by himself than with the others. So, 0.5 days is not a reasonable answer. Joe can paint a house by himself in about 7.2 days. Answer the question. 2002, Renaissance Learning, Inc.

24 II ObJectlv_Provo quadratic formula You are going to demonstrate the steps in the proof of the quadratic formula using the equation 2X2-3x + 1 = o. For the step in the proof just before taking the square root of each side, what number would appear on the side with no variables? You are going to demonstrate the steps in the proof of the quadratic formula using the equation 8x 2 + 9x - 5 = O. For the step in the proof just before taking the square root of each side, what number would appear on the side with no variables? Review proof of the quadratic formula. You can prove that the quadratic formula works by completing the square. If you start with the general formula, ar + bx + c = 0, you end up with the quadratic formula. Move the constant to the 2r-3x+1-1=0-1 right side of the equal sign. 2r-3x= -1 Y 3x - 1 Divide all terms by 2-2"=2 the coefficient on r. _2 3x _ 1 X Find the square of half the coefficient on x. Add this to both sides of the equation. The coefficient on x is ~. r - 3x = -" ) 3x 9 1 r = 16 (~ ~r=(~r=196 3)2 1 Factor the left side. (x -"4 = 16 Review proof of the quadratic formula. Move the constant to the right side of the equal sign. Divide all terms by the coefficient on r. Find the square of half the coefficient on x. Add this to both sides of the equation. The coefficient on x is Half the coefficient is Half the coefficient squared is Factor the left side. Step 6: Answer the question. To finish solving for x, you would take the square root of both sides of the equation. We have stopped just before that step to answer the question. The number that appears on the side with no variables is 1~. Step 6: Answer the question. The number that appears on the side with no variables IS ~ Renaissance Learning. Inc.

Algebra I. Slide 1 / 175. Slide 2 / 175. Slide 3 / 175. Quadratics. Table of Contents Key Terms

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