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EXAMPLE 1 PowerPoint Presentation

EXAMPLE 1

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EXAMPLE 1

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  1. x = 2a –b+b2– 4ac –3+32– 4(1)(–2) x = 2(1) 17 17 –3+ –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.

  2. 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. 

  3. 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.

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

  5. –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.

  6. ? –(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 = 2x2 + 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.

  7. 5+i 115 10 1 2 for Examples 1, 2, and 3 GUIDED PRACTICE Use the quadratic formula to solve the equation. x2= 6x – 4 ANSWER 3 + 5 4x2 – 10x = 2x – 9 ANSWER 1 7x – 5x2 – 4 = 2x + 3 ANSWER