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1. Write 15 x 2 + 6 x = 14 x 2 – 12 in standard form.

Warm Up #4. 1. Write 15 x 2 + 6 x = 14 x 2 – 12 in standard form. ANSWER. x 2 + 6 x +12 = 0. 2. Evaluate b 2 – 4 ac when a = 3, b = –6, and c = 5. –24. ANSWER. Classwork Answers…. 4.8 (13-27 odd). 4.6 (multiples of 3).

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1. Write 15 x 2 + 6 x = 14 x 2 – 12 in standard form.

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  1. Warm Up #4 1. Write15x2 + 6x = 14x2 – 12in standard form. ANSWER x2 + 6x +12 = 0 2. Evaluateb2 – 4acwhena = 3, b = –6, andc = 5. –24 ANSWER

  2. Classwork Answers… 4.8 (13-27 odd)

  3. 4.6 (multiples of 3)

  4. When solving quadratic equations, we’re looking for the x values where the graph crosses the x axis One of the methods we use to solve quadratic equations is called the Quadratic Formula Using the a, b, and c from ax2 + bx + c = 0 Must be equal to zero

  5. x = 2a –b+b2– 4ac –3+32– 4(1)(–2) x = 2(1) 17 –3+ x = 2 EXAMPLE 1 Solve an equation with two real solutions Solvex2 + 3x = 2. x2 + 3x = 2 Write original equation. x2 + 3x – 2 = 0 Write in standard form. Quadratic formula a = 1, b = 3, c = –2 Simplify. or

  6. 30+(–30)2– 4(25)(9) x = 30+ 0 2(25) x = 50 3 x = 5 3 ANSWER 5 The solution is EXAMPLE 2 Solve an equation with one real solutions Solve25x2 – 18x = 12x – 9. 25x2 – 18x = 12x – 9. Write original equation. 25x2 – 30x + 9 = 0. Write in standard form. a = 25, b = –30, c = 9 Simplify. Simplify.

  7. –4 +42 – 4(–1)(–5) x = –4 +–4 2(–1) x = –2 –4 + 2i x = –2 ANSWER The solution is 2 + i and 2 – i. EXAMPLE 3 Solve an equation with imaginary solutions Solve–x2 + 4x = 5. –x2 + 4x = 5 Write original equation. –x2 + 4x – 5 = 0. Write in standard form. a = –1, b = 4, c = –5 Simplify. Rewrite using the imaginary unit i. x = 2 +i Simplify.

  8. x = 2a –b+b2– 4ac for Examples 1, 2, and 3 GUIDED PRACTICE Use the quadratic formula to solve the equation. x2= 6x – 4 x2– 6x + 4 = 0 a = 1 b = -6 c = 4

  9. x = 2a –b+b2– 4ac for Examples 1, 2, and 3 GUIDED PRACTICE Use the quadratic formula to solve the equation. 4x2 – 10x = 2x – 9 4x2– 12x + 9 = 0 a = 4 b = -12 c = 9

  10. EXAMPLE 4 Use the discriminant If the quadratic equation is in the standard form ax2 + bx + c = 0 The discriminant can be found using b2 – 4ac If b2 – 4ac < 0 There are Two Imaginary solutions If b2 – 4ac = 0 There is One Real solution If b2 – 4ac > 0 There are Two Real solutions Discriminant Equation b2 – 4ac Solution(s) ( –8)2 – 4(1)(17) = –4 Two imaginary a. x2 – 8x + 17 = 0 (–8)2 – 4(1)(16) = 0 One real b. x2 – 8x + 16 = 0 (–8)2 – 4(1)(15) = 4 b. x2 – 8x + 15 = 0 Two real

  11. for Example 4 GUIDED PRACTICE Find the discriminant of the quadratic equation and give the number and type of solutions of the equation. 3x2 + 12x + 12 = 0 2x2 + 4x – 4 = 0 b2 – 4ac b2 – 4ac (4)2 – 4(2)(-4) = 48 (12)2 – 4(3)(12) = 0 Two Real solutions One Real solution 8x2 = 9x – 11 7. 4x2 + 3x + 12 = 3 – 3x 8x2 – 9x + 11 = 0 4x2 + 6x + 9 = 0 b2 – 4ac b2 – 4ac (6)2 – 4(4)(9) = -108 (-9)2 – 4(8)(11) = -271 Two Imaginary solutions Two Imaginary solutions

  12. Classwork Assignment: WS 4.8 (1-27 odd)

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