Location via proxy:   [ UP ]  
[Report a bug]   [Manage cookies]                
SlideShare a Scribd company logo

10.7

•Download as PPTX, PDF•
•

Here are the steps to solve the quadratic equations using the quadratic formula: 1) Write the equation in standard form: ax^2 + bx + c = 0 2) Identify the coefficients a, b, c 3) Plug the coefficients into the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a 4) Simplify the solution The quadratic formula can be used to find the roots (solutions) of any quadratic equation in standard form. Other solution methods like factoring or completing the square may also work for certain equations.

1 of 16
10.7
Derive the Quadratic Formula  ax 2 + bx + c = 0
The Quadratic Formula  The roots of the polynomial ax 2 + bx + c and the solutions of the quadratic equation ax 2 + bx + c = 0 -b ± b2 - 4ac are x = where a ¹ 0 and b2 - 4ac ³ 0 2a
Example 1 Find the roots of a polynomial Find the roots of x2 – 6x + 3. SOLUTION The roots of x2 – 6x + 3 are the values of x for which x2 – 6x + 3 = 0. –b + – b2 – 4ac x= Quadratic formula 2a – ( – 6) + – ( – 6)2 – 4( 1 )( 3 ) Substitute values in the quadratic x= formula: a = 1, b = – 6, and c = 3. 2( 1 )
Example 1 Find the roots of a polynomial 6 + 24 = – Simplify. 2 6 +2 6 = – Simplify radical. 2 2(3 + 6 ) – = = 3 + – 6 Divide out factor of 2. 2 ANSWER The roots of x2 – 6x + 3 are 3 + 6 and 3 – 6.
Example 1 Find the roots of a polynomial CHECK Substitute each root for x. The polynomial should simplify to 0. (3 + 6 )2 – 6 (3 + 6) + 3 = 9 + 6 6 + 6 – 18 – 6 6 + 3 = 0 (3 – 6 )2 – 6 (3 – 6) + 3 = 9 – 6 6 + 6 – 18 + 6 6 + 3 = 0
Example 2 Multiple Choice Practice Which is one of the solutions to the equation 2x2 – 7 = x? 1 1 – 57 + 57 4 4 –1 + 57 1+ 57 4 4 SOLUTION Write original 2x2 – 7 = x equation. Write in standard 2x2 – x – 7 = 0 form.
Example 2 Multiple Choice Practice –b+ – b2 – 4ac Quadratic formula x = 2a Substitute values – ( – 1) + ( – 1)2 – 4( 2 )(–7) in the quadratic – formula: = 2( 2 ) a = 2, b = –1, and c = –7. 1 + 57 – = Simplify. 4 ANSWER 1 + 57 One solution is . 4 The correct answer is D.
Methods for Solving Quadratic Equations
Example 4 Choose a solution method Tell what method(s) you would use to solve the quadratic equation. Explain your choice(s). a. 10x2 – 7 = 0 b. x2 + 4x = 0 c. 5x2 + 9x – 4 = 0 SOLUTION a. The quadratic equation can be solved using square roots because the equation can be written in the form x2 d. =
Example 4 Choose a solution method b. The quadratic equation can be solved by factoring because the expression x2 + can be factored easily. 4x Also, the equation can be solved by completing the square because the equation is of the form where a ax2and b is an=even number. 1 + bx + c 0 = c. The quadratic equation cannot be factored easily, and completing the square will result in many fractions. So, the equation should be solved using the quadratic formula.
Example 3 Use the quadratic formula FILM PRODUCTION For the period 1971 – 2001, the number y of films produced in the world can be modeled by the function y = 10x2 – 94x + 3900 where x is the number of years since 1971. In what year were 4200 films produced? SOLUTION y = 10x2 – 94x + 3900 Write function. 4200 = 10x2 – 94x + 3900 Substitute 4200 for y. 0 = 10x2 – 94x – 300 Write in standard form.
Example 3 Use the quadratic formula Substitute values in the – ( – 94) + – ( – 94)2 – 4 (10)(–300) quadratic formula: a = 10, x = 2(10) b = –94, and c = –300. 94 + 20,836 – Simplify. = 20 94 + 20,836 The solutions of the equation are ≈ 12 and 20 94 – 20,836 ≈ –3. 20 ANSWER There were 4200 films produced about 12 years after 1971, or in 1983.
10.7 Warm-up (Day 1) Use the quadratic formula to Find the roots 1. x 2 + 4x -1 Use the quadratic formula to solve the equation 2. x 2 - 8x +16 = 0 3. x 2 - 5x = -21 4. 4z 2 = 7z + 2
10.7 Warm-up (Day 2) Use the quadratic formula to solve the equation 1. n2 +1 = 5n 2. 2z + 4 = 3z 2
10.7 Warm-up (Day 3) Use the quadratic formula to solve the equation 1. 4x 2 + 6x = 3x 2 - 4x +1 2. 7r 2 - 2r = 6

More Related Content

10.7

  • 2. Derive the Quadratic Formula  ax 2 + bx + c = 0
  • 3. The Quadratic Formula  The roots of the polynomial ax 2 + bx + c and the solutions of the quadratic equation ax 2 + bx + c = 0 -b ± b2 - 4ac are x = where a ¹ 0 and b2 - 4ac ³ 0 2a
  • 4. Example 1 Find the roots of a polynomial Find the roots of x2 – 6x + 3. SOLUTION The roots of x2 – 6x + 3 are the values of x for which x2 – 6x + 3 = 0. –b + – b2 – 4ac x= Quadratic formula 2a – ( – 6) + – ( – 6)2 – 4( 1 )( 3 ) Substitute values in the quadratic x= formula: a = 1, b = – 6, and c = 3. 2( 1 )
  • 5. Example 1 Find the roots of a polynomial 6 + 24 = – Simplify. 2 6 +2 6 = – Simplify radical. 2 2(3 + 6 ) – = = 3 + – 6 Divide out factor of 2. 2 ANSWER The roots of x2 – 6x + 3 are 3 + 6 and 3 – 6.
  • 6. Example 1 Find the roots of a polynomial CHECK Substitute each root for x. The polynomial should simplify to 0. (3 + 6 )2 – 6 (3 + 6) + 3 = 9 + 6 6 + 6 – 18 – 6 6 + 3 = 0 (3 – 6 )2 – 6 (3 – 6) + 3 = 9 – 6 6 + 6 – 18 + 6 6 + 3 = 0
  • 7. Example 2 Multiple Choice Practice Which is one of the solutions to the equation 2x2 – 7 = x? 1 1 – 57 + 57 4 4 –1 + 57 1+ 57 4 4 SOLUTION Write original 2x2 – 7 = x equation. Write in standard 2x2 – x – 7 = 0 form.
  • 8. Example 2 Multiple Choice Practice –b+ – b2 – 4ac Quadratic formula x = 2a Substitute values – ( – 1) + ( – 1)2 – 4( 2 )(–7) in the quadratic – formula: = 2( 2 ) a = 2, b = –1, and c = –7. 1 + 57 – = Simplify. 4 ANSWER 1 + 57 One solution is . 4 The correct answer is D.
  • 9. Methods for Solving Quadratic Equations
  • 10. Example 4 Choose a solution method Tell what method(s) you would use to solve the quadratic equation. Explain your choice(s). a. 10x2 – 7 = 0 b. x2 + 4x = 0 c. 5x2 + 9x – 4 = 0 SOLUTION a. The quadratic equation can be solved using square roots because the equation can be written in the form x2 d. =
  • 11. Example 4 Choose a solution method b. The quadratic equation can be solved by factoring because the expression x2 + can be factored easily. 4x Also, the equation can be solved by completing the square because the equation is of the form where a ax2and b is an=even number. 1 + bx + c 0 = c. The quadratic equation cannot be factored easily, and completing the square will result in many fractions. So, the equation should be solved using the quadratic formula.
  • 12. Example 3 Use the quadratic formula FILM PRODUCTION For the period 1971 – 2001, the number y of films produced in the world can be modeled by the function y = 10x2 – 94x + 3900 where x is the number of years since 1971. In what year were 4200 films produced? SOLUTION y = 10x2 – 94x + 3900 Write function. 4200 = 10x2 – 94x + 3900 Substitute 4200 for y. 0 = 10x2 – 94x – 300 Write in standard form.
  • 13. Example 3 Use the quadratic formula Substitute values in the – ( – 94) + – ( – 94)2 – 4 (10)(–300) quadratic formula: a = 10, x = 2(10) b = –94, and c = –300. 94 + 20,836 – Simplify. = 20 94 + 20,836 The solutions of the equation are ≈ 12 and 20 94 – 20,836 ≈ –3. 20 ANSWER There were 4200 films produced about 12 years after 1971, or in 1983.
  • 14. 10.7 Warm-up (Day 1) Use the quadratic formula to Find the roots 1. x 2 + 4x -1 Use the quadratic formula to solve the equation 2. x 2 - 8x +16 = 0 3. x 2 - 5x = -21 4. 4z 2 = 7z + 2
  • 15. 10.7 Warm-up (Day 2) Use the quadratic formula to solve the equation 1. n2 +1 = 5n 2. 2z + 4 = 3z 2
  • 16. 10.7 Warm-up (Day 3) Use the quadratic formula to solve the equation 1. 4x 2 + 6x = 3x 2 - 4x +1 2. 7r 2 - 2r = 6

Editor's Notes

  1. Day 2
  2. 1. -2+/-rad(5) 2. 4 3. no solution 4. -1/4, 2
  3. 1. (5+/-rad(21))/2 2. (1+/-rad(13))/3
  4. 1. -5 +/- rad(26) 2. (1 +/- rad(43)) / 7