Solutions Algebra: 8.2.2 Solving Quadratic Equations Name ______________________________ Block _____ Date ______ Bell Work 1/15 Factor each quadratic equation completely. Show sufficient work. a. y = 2x2 + 8x + 8 2 π¦ = 2(π₯ + 4π₯ + 4) π¦ = π( π + π) π 2 2π₯ π₯ π₯2 π₯ 4 b. y = 25x2 β 64 π¦ = (ππ + π)(ππ β π) 2π₯ 2 8-64. Graph y = x β 2x β 15 without using the calculator. (Adjust scale as needed) 2 3 3π₯ β15 π₯ π₯ 2 π₯ a. Factored form: β5π₯ β5 b. Find the roots: Explain how to find. c. Find the y-intercept. Explain how to find. π¦ = (π₯ + 3)(π₯ β 5) π₯+3=0 π₯β5=0 π₯ = β3 π₯=5 Set Factors equal to 0 and solve -3 π¦ = β15 It is the last number in the standard equation d. Find the x-coordinate of the vertex. Explain how to find. β3 + 5 =1 2 Average the roots 5 -15 (1,-16) π¦ = (1 + 3)(1 β 5) π¦ = (4)(β4) π¦ = β16 Plug x into an equation and solve 2 8-65. Graph y = 2x + 5x β 12 without using the calculator. e. Find the y-coordinate of the vertex. Explain how to find. 4 8π₯ β12 π₯ 2π₯ 2 β3π₯ a. Factored form: 2π₯ β3 b. Find the roots: π = (π + π)(ππ β π) π₯+4 =0 π = βπ 2π₯ β 3 = 0 2π₯ = 3 π = π. π c. Find the y-intercept. π = βππ d. Find the x-coordinate of the vertex. β4 + 1.5 π₯= = β1.25 2 e. Find the y-coordinate of the vertex. π¦ = (β1.25 + 4)(2(β1.25) β 3) π¦ = (2.25)(β5.5) = β15.25 -4 1.5 -12 (-1.25, -15.25) 8-68. Graph y = x2 β 2x β 8 without using the calculator. 2 2π₯ β8 π₯ π₯ 2 β4π₯ a. Factored form: π₯ β4 b. Find the roots: π = (π + π)(π β π) π = βπ c. Find the y-intercept. π = βπ π=π -2 d. Find the x-coordinate of the vertex. βπ + π π= =π π e. Find the y-coordinate of the vertex. π = ππ β π(π) β π = βπ 4 -8 (1,-9) 8-69. Graph y = x2 + x β 6 without using the calculator. 3 3π₯ β6 π₯ π₯ 2 β2π₯ a. π₯ Factored form: β2 b. Find the roots: π = (π + π)(π β π) π = βπ π=π π = βπ c. Find the y-intercept. d. Find the x-coordinate of the vertex. βπ + π π= = βπ. π Find the y-coordinate of the vertex. π e. π = (βπ. π + π)(βπ. π β π) π = (π. π)(βπ. π) π = βπ. ππ 8-70. Compare the two equations below. (x + 2)(x β 1) = 0 and (x + 2) + (x β 1) = 0 a. How are the equations different? The first equation is multiplying binomials (quadratic). The second equation is adding binomials (linear) b. Solve both equations. (π₯ + 2)(π₯ β 2) = 0 π₯+2 =0 π₯β1 =0 π = βπ π=π 2 Solutions π₯+2+π₯β1= 0 2π₯ + 1 = 0 2π₯ = β1 βπ π= π 1 Solution The following are answers only. Show SUFFICIENT WORK!. 8-71. For each equation below, solve for x. a. (x β 2)(x + 8) = 0 b. (3x β 9)(x β 1) = 0 c. (x + 10)(2x β 5) = 0 π = βππ (x β 7)2 = 0 d. π=π π = βπ π=π π=π π=π π= π π 8-72. Solve the system of equations. 5x β 2y = 4 x=0 (π, βπ) 8-73. The (x-intercepts) of the graph of y = 2x2 β 16x + 30 are (3, 0) and (5, 0). a. b. What is the x-coordinate of the vertex? How do you know? 4, it is halfway between 3 and 5. Use your answer to part (a) above to find the y-coordinate of the vertex. Then write the vertex as a point (x, y). π¦ = 2(4)2 β 16(4) + 30 = β2 8-74. Factor each quadratic below completely. a. 2x2 β 2x β 4 b. 4x2 β 24x + 36 = π(π β π)π = π(π β π)(π + π) EOC Practice: βπ π ππ ππ (π, βπ)
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