Quadratic Equations Quiz 1 Key Name _________________________________________ Period _____ Date ________ Directions: Read each question fully and carefully. Answer each question completely. You must show your work. Each question is worth five (5) points. 1. Solve by completing the square: x 2 ο 6x ο« 2 ο½ 0 π₯ 2 β 6π₯ = β2 π₯ 2 β 6π₯ + 9 = 7 π₯β3 2 =7 π₯β3=± 7 π₯ = 3± 7 Subtract 2 Add β6/2 2 = 9 Write as perfect square Square root Solution 2. Solve using the quadratic formula: 4 x2 ο« 2 x ο« 1 ο½ 0 π=4 π=2 π₯= π₯= ;π± π 2 ;4ππ 2π ;2± 4;16 8 1 π₯ = β4 ± 3 4 π=1 = = ;2± 22 ;4β4β1 Quadratic Formula 2β4 ;2± ;12 8 = ;2±π 4 3 8 π Solution 3. Solve the following problem. Approximate answers involving radicals to the nearest hundredth. Find the dimensions of a rectangle whose perimeter is 10 cm and whose area is 3 cm2. πππππππ‘ππ = 2π€ + 2β = 10 ππ ππππ = π€ β β = 3 ππ2 β =5βπ€ solve perimeter equation for h h π€ 5βπ€ =3 substitute into area equation 2 5π€ β π€ = 3 simplify w π€ 2 β 5π€ + 3 = 0 standard form π=1 π = β5 π=3 π€= π€= ; ;5 ± ;5 2 ;4β1β3 Quadratic Formula 2β1 5 ± 25 β 12 5 ± 13 = 2 2 Dimensions: 0.70 cm ο΄ 4.30 cm -1- Quadratic Equations Quiz 1 Key Name _________________________________________ Period _____ Date ________ Directions: Read each question fully and carefully. Answer each question completely. You must show your work. Each question is worth five (5) points. 1. Solve by completing the square: x 2 ο 5x ο½ 3 π₯ 2 β 5π₯ + 5 2 π₯β2 25 = 5 37 Add β5/2 4 2 = 25 4 Write as perfect square 4 37 π₯β2=± π₯= = 4 37 Square root 4 5± 37 Solution 2 2. Solve using the quadratic formula: 3x 2 ο« 9 x ο½ ο2 3π₯ 2 + 9π₯ + 2 = 0 π=3 π₯= π₯= Add 2 π=9 ;π± π 2 ;4ππ 2π ;9± 81;24 6 π=2 = = ;9± 92 ;4β3β2 Quadratic Formula 2β3 ;9± 57 Solution 6 3. Solve the following problem. Approximate answers involving radicals to the nearest hundredth. A sidewalk of uniform width has area 180 ft2 and surrounds a flower bed that is 11 ft wide and 13 ft long. Find the width of the sidewalk. π΄πππ ππ πππππ€πππ = 2 π΄πππ ππ π΅ππ‘π‘ππ + 2 π΄πππ ππ ππππ w 13ο’ 11ο’ Bottom Side π΄ = 2 β 11π€ + 2 β 13 + 2π€ π€ = 180 Area of Sidewalk 22π€ + 26π€ + 4π€ 2 = 180 4π€ 2 + 48π€ β 180 = 0 Standard Form π€ 2 + 12π€ β 45 = 0 Simplify π€= ;12± 122 ;4β1β ;45 Quadratic Formula 2β1 π€= ;12± 144:180 π€= ;12±18 2 2 = ;12± 324 = β15, 3 π€ = 3 ππ‘ 2 Solution Width cannot be negative -2- Quadratic Equations Quiz 1 Key Name _________________________________________ Period _____ Date ________ Directions: Read each question fully and carefully. Answer each question completely. You must show your work. Each question is worth five (5) points. 1. Solve by completing the square: x2 ο 8x ο« 3 ο½ 0 π₯ 2 β 8π₯ = β3 π₯ 2 β 8π₯ + 16 = 13 π₯ β 4 2 = 13 π₯ β 4 = ± 13 π₯ = 4 ± 13 Subtract 3 Add β8/2 2 = 16 Write as perfect square Square root Solution 2. Solve using the quadratic formula: 3x 2 ο« 3x ο« 1 ο½ 0 π=3 π=3 π₯= π₯= ;π± π 2 ;4ππ 2π ;3± 9;12 6 1 π₯ = β2 ± 3 6 = = π=1 ;3± 32 ;4β3β1 Quadratic Formula 2β3 ;3± ;3 6 = ;3±π 3 6 π Solution 3. Solve the following problem. Approximate answers involving radicals to the nearest hundredth. Find the dimensions of a rectangle whose perimeter is 42 cm and whose area is 20 cm2. h w πππππππ‘ππ = 2π€ + 2β = 42 ππ ππππ = π€ β β = 20 ππ2 β = 21 β π€ π€ 21 β π€ = 20 21π€ β π€ 2 = 20 π€ 2 β 21π€ + 20 = 0 π=1 π = β21 π = 20 π€= π€= ; ;21 ± ;21 2 ;4β1β20 2β1 solve perimeter equation for h substitute into area equation simplify standard form Quadratic Formula 21 ± 441 β 80 21 ± 361 21 ± 19 = = 2 2 2 Dimensions: 1.00 cm ο΄ 20.00 cm -3- Quadratic Equations Quiz 1 Key Name _________________________________________ Period _____ Date ________ Directions: Read each question fully and carefully. Answer each question completely. You must show your work. Each question is worth five (5) points. 1. Solve by completing the square: x2 ο« 5x ο½ 3 25 π₯ 2 + 5π₯ + 5 2 π₯+2 = 5 37 Add 5/2 4 2 = 25 4 Write as perfect square 4 37 π₯+2=± π₯= = 4 37 Square root 4 ;5± 37 Solution 2 2. Solve using the quadratic formula: 4 x 2 ο 7 x ο½ ο2 4π₯ 2 β 7π₯ + 2 = 0 π=4 π₯= π₯= Add 2 π = β7 ;π± π 2 ;4ππ 2π 7± 49;32 8 = = π=2 ;7 2 ;4β4β2 7± Quadratic Formula 2β4 7± 17 Solution 8 3. Solve the following problem. Approximate answers involving radicals to the nearest hundredth. A sidewalk of uniform width has area 352 ft2 and surrounds a flower bed that is 17 ft wide and 19 ft long. Find the width of the sidewalk. π΄πππ ππ πππππ€πππ = 2 π΄πππ ππ π΅ππ‘π‘ππ + 2 π΄πππ ππ ππππ w 19ο’ 17ο’ Bottom Side π΄ = 2 β 17π€ + 2 β 19 + 2π€ π€ = 352 Area of Sidewalk 34π€ + 38π€ + 4π€ 2 = 352 4π€ 2 + 72π€ β 352 = 0 Standard Form π€ 2 + 18π€ β 88 = 0 Simplify π€= ;18± 182 ;4β1β ;88 Quadratic Formula 2β1 π€= ;18± 324:352 π€= ;18±26 2 2 = ;18± 676 = β22, 4 π€ = 4 ππ‘ 2 Solution Width cannot be negative -4-
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