1 / 13

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).

cato
Download Presentation

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

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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)

More Related