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

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. _. +. 3 . A student is solving an equation by completing the square. Write the step in the solution that appears

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

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

  2. _ + 3. A student is solving an equation by completing the square. Write the step in the solution that appears just before “(x – 3) = 5.” (x – 3)2 = 25 ANSWER

  3. x = 2a –b+b2– 4ac –3+32– 4(1)(–2) x = 2(1) 17 –3+ 17 –3+ x = x = 2 2 ANSWER The solutions are 0.56and 17 –3– x = –3.56. 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.

  4. EXAMPLE 1 Solve an equation with two real solutions CHECK Graph y = x2 + 3x – 2 and note that the x-intercepts are about 0.56 and about –3.56. 

  5. 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 solution 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. .

  6. EXAMPLE 2 Solve an equation with one real solution CHECK Graph y = –5x2 – 30x + 9 and note that the only x-intercept is 0.6 = . 3  5

  7. – 4+ 42 – 4(– 1)(– 5) x = – 4+– 4 2(– 1) x = – 2 – 4+ 2i x = – 2 ANSWER The solutions are 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. ? –(2 + i)2 + 4(2 + i) = 5 ? –3 – 4i + 8 + 4i = 5 5 = 5  EXAMPLE 3 Solve an equation with imaginary solutions CHECK Graph y = −x2 + 4x – 5. There are no x-intercepts. So, the original equation has no real solutions. The algebraic check for the imaginary solution 2 + iis shown.

  9. The solutions are 3 – 5 3 + 5 and The solution is . 3 2 for Examples 1, 2, and 3 GUIDED PRACTICE Use the quadratic formula to solve the equation. x2= 6x – 4 ANSWER . 4x2 – 10x = 2x – 9 ANSWER

  10. ANSWER 5+ i 5–i 115 115 The solutions are and . 10 10 for Examples 1, 2, and 3 GUIDED PRACTICE Use the quadratic formula to solve the equation. 7x – 5x2 – 4 = 2x + 3

  11. – b+b2– 4ac x = 2a EXAMPLE 4 Use the discriminant Find the discriminant of the quadratic equation and give the number and type of solutions of the equation. b.x2 – 8x + 16 = 0 c.x2 – 8x + 15 = 0 a.x2 – 8x + 17 = 0 SOLUTION Discriminant Solution(s) Equation ax2 + bx + c = 0 b2 – 4ac a. x2 – 8x + 17 = 0 (–8)2 – 4(1)(17) = – 4 Two imaginary:4 +i b. x2 – 8x + 16 = 0 (–8)2 – 4(1)(16) = 0 One real: 4 c. x2 – 8x + 15 = 0 (–8)2 – 4(1)(15) = 4 Two real: 3, 5

  12. for Example 4 GUIDED PRACTICE Find the discriminant of the quadratic equation and give the number and type of solutions of the equation. 2x2 + 4x – 4 = 0 ANSWER 48; Two real solutions 3x2 + 12x + 12 = 0 0; One real solution ANSWER 8x2 = 9x – 11 – 271; Two imaginary solutions ANSWER

  13. for Example 4 GUIDED PRACTICE Find the discriminant of the quadratic equation and give the number and type of solutions of the equation. 7x2 – 2x = 5 144; Two real solutions ANSWER 4x2 + 3x + 12 = 3 – 3x – 108; Two imaginary solutions ANSWER 3x – 5x2 + 1 = 6 – 7x 0; One real solution ANSWER

  14. EXAMPLE 5 Solve a vertical motion problem Juggling A juggler tosses a ball into the air. The ball leaves the juggler’s hand 4feet above the ground and has an initial vertical velocity of 40feet per second. The juggler catches the ball when it falls back to a height of 3feet. How long is the ball in the air? SOLUTION Because the ball is thrown, use the model h = –16t2 + v0t +h0. To find how long the ball is in the air, solve for twhen h = 3.

  15. – 40+ 402– 4(– 16)(1) t = 2(– 16) – 40+ 1664 t = – 32 t – 0.025 or t 2.5 EXAMPLE 5 Solve a vertical motion problem h= –16t2 + v0t + h0 Write height model. 3= –16t2 + 40t + 4 Substitute 3 for h, 40 for v0, and 4 for h0. 0 = –16t2+ 40t + 1 Write in standard form. Quadratic formula Simplify. Use a calculator.

  16. EXAMPLE 5 Solve a vertical motion problem ANSWER Reject the solution – 0.025 because the ball’s time in the air cannot be negative. So, the ball is in the air for about 2.5seconds.

  17. for Example 5 GUIDED PRACTICE WHAT IF?In Example 5, suppose the ball leaves the juggler’s hand with an initial vertical velocity of 50feet per second. How long is the ball in the air? ANSWER The ball is in the air for about 3.1seconds.

  18. 1 2 ANSWER 4 + 2i,4 – 2i ANSWER – ANSWER 3 ± 11 Daily Homework Quiz Use the quadratic formula to solve the equation. 1. 4x2 – 4x + 1 = 0 2. – x2 +8x = 20 3. – x2 =6x – 2

  19. 57; two real solutions ANSWER about 2.6 sec ANSWER Daily Homework Quiz 4. Find the discriminant of 2x2 + 3x – 6 = 0 and give the number and type of solutions of the equation. 5. An object is thrown upward from a height of 15 feet at an initial velocity of 35 feet per second. How long will it take for the object to hit the ground?

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