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Algebra 2

Algebra 2. Section 4.7 Complete the Square Section 4.8 Quadratic Formula and the Discriminant. 4.7 Complete the Square . Standard Form ax 2 + bx + c Intercept Form m(x - p)(x - q) Vertex Form m(x -d) 2 + b. Example 1: 4.7.

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Algebra 2

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  1. Algebra 2 Section 4.7 Complete the Square Section 4.8 Quadratic Formula and the Discriminant

  2. 4.7 Complete the Square • Standard Form ax2 + bx + c • Intercept Form m(x - p)(x - q) • Vertex Form m(x -d)2 + b

  3. Example 1: 4.7 • Write y = x2 + 14x + 44 in vertex form. Then identify the vertex. Step 1: Complete the square (find c) 14/2 = 7 7*7 = 49 = c Step 2: Add c to both sides of the equation. y + 49 = x2 + 14x + 49 + 44 Step 3: Factor only your perfect square trinomial y + 49 = (x + 7)2 + 44 Step 4: Solve for y. y = (x + 7)2 + 44 - 49 y = (x + 7)2 - 5 (-7, -5)

  4. Assignment: Page 288 #36 • Find the value of x. Area of a parallelogram = 48 A = bh 48 = (x + 6)x 48 = x2 + 6x Step 1: complete the square 48 + 9 = x2 + 6x + 9 Step 2: factor and solve 57 = (x + 3)2 x x + 6

  5. Assignment: Page 289 #44 Write the quadratic function in vertex form. Then identify the vertex. y = x2 + 20x + 90 Step 1: complete the square 20/2 = 10, 10*10 = 100 = c Step 2: add c to both sides y + 100 = x2 + 20x + 100 + 90 Step 3: factor only perfect square trinomial y + 100 = (x + 10)2 + 90 Step 4: solve for y y = (x + 10)2 + 90 - 100 y = (x + 10)2 - 10 Vertex: (-10,-10)

  6. Assignment: Page 289 # 48 - 52 eoe You have 5 minutes!

  7. 4.8 The Quadratic Formula and the Discriminant • Standard Form: ax2 + bx + c • Quadratic Formula: formula that solves any quadratic equation • Discriminant: b2 - 4ac

  8. Discriminant: will tell you the number of solutions • When b2 - 4ac > 0, then the equation has two solutions and the graph has 2 x-intercepts • When b2 - 4ac = 0, then the equation has one solution and the graph has 1 x-intercept • When b2 - 4ac < 0, then the equation has two imaginary solutions and the graph has no x-intercepts

  9. two solutions Example 1: • Solve x2 + 7x = 6 Step 1: put equation in standard form x2 + 7x = 6 -6 -6 x2 + 7x - 6 = 0 Step 2: label a, b, and c a = 1, b = 7, c = -6 Step 3: Plug and Chug

  10. one solution Example 2: • Solve 2x2 - 8x + 8 = 0 Step 1: Standard form Already in it! Step 2: label a, b, and c a = 2, b = -8, c = 8 Step 3: Plug and Chug

  11. imaginary solutions Example 3: • Solve -x2 + 2x = 5 Step1: standard form -x2 + 2x = 5 -5 -5 -x2 + 2x - 5 = 0 Step 2: label a, b, and c a = -1, b = 2, c = -5 Step 3: Plug and Chug

  12. Assignment: Page 288 and 296 • # 36, 44 - 52 eoe • # 4 - 60 eoe

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