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Solving quadratic equations

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Asawari Warkad
Asawari Warkad

The document discusses solving quadratic equations by factoring. It begins with an example problem and explanation of the zero factor property. It then outlines the steps to solve quadratic equations by factoring: 1) write the equation in standard form, 2) factor completely, 3) set each factor equal to 0, 4) solve each equation, and 5) check solutions. Several example problems are worked through demonstrating this process. The document concludes with examples of using factoring to solve real-world problems involving quadratic equations.

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Equations and Quadratic functions Objective Solving quadratic equation by Factorisation
Warm up • Solve the given equation.     y y y y y    3 6 1 2 6 3 11 5 3 5 11 3
Explanation
11.6 – Solving Quadratic Equations by Factoring A quadratic equation is written in the Standard Form, ax2 bx  c  0 where a, b, and c are real numbers and . 0 a  Examples: 2 x 7x 12  0 xx 7  0 2 3x  4x 15 (standard form)
11.6 – Solving Quadratic Equations by Factoring Zero Factor Property: If a and b are real numbers and if , 0 ab  then or . 0 a  0 b  Examples: xx 7  0 x  0 x 7  0 x  0 x  7
11.6 – Solving Quadratic Equations by Factoring Zero Factor Property: If a and b are real numbers and if , 0 ab  then or . 0 a  0 b  Examples: x103x6  0 x 10  0 3x6  0 x 1010  010 3x 66  06 x 10 x  3 6 3 3 3x  6 x  2
11.6 – Solving Quadratic Equations by Factoring Solving Quadratic Equations: 1) Write the equation in standard form. 2) Factor the equation completely. 3) Set each factor equal to 0. 4) Solve each equation. 5) Check the solutions (in original equation).
11.6 – Solving Quadratic Equations by Factoring 2 x 3x 18 x2 3x 18  0 Factors of 18: 1, 18 2, 9 3, 6 x3 x 3  0 x 6  0 x  3 x  6     2 6 3 6 18 3618 18 18 18      2 3 3 3 18 99 18 18 18   6 x 0
11.6 – Solving Quadratic Equations by Factoring If the Zero Factor Property is not used, then the solutions will be incorrect 2 x 3x 18 xx3 18 x 18     2 18 3 18 18 32454 18 270 18      2 21 3 21 18 4416318 378 18 x318 x33183 x  21
11.6 – Solving Quadratic Equations by Factoring xx 4  5 x2  4x  5 2 x  4x 5  0 x 1x 5  0 x1 0 x 5  0 x  1 x  5
11.6 – Solving Quadratic Equations by Factoring x 33x 2  0 3x2  7x  6 30x 3 2 0 x  3 x  2 3 x  x3x 7  6 2 3 7 6 0 x x    32x  Factors of 3: 1, 3 Factors of 6: 1, 6 2, 3
11.6 – Solving Quadratic Equations by Factoring 9x2  24x  16 2 9x 24x 16  0 9 and 16 are perfect squares 3x 43x 4  0 3x4  0 3x  4 4 3 x 
11.6 – Solving Quadratic Equations by Factoring 2x3 18x  0 2x 2x   2 9 x  0    3 x    3 x  0  x 3  0 x3 0 2x  0 x  0 x  3 x  3
11.6 – Solving Quadratic Equations by Factoring x 33x2 20x 7  0 3: Factors of 1, 37 : Factors of 1, 7 x3 30 x    7 x  0   3 1 x  70x 3 1 0 x  x  3 x  7 3x  1 1 3 x  
Continuous assessment • Q1: Solve these equations by factorisation. a) (x-2)(2x+1) = 0 b) 3x2 – 27x = 0 c) 2x2 – 7x + 6 = 0 Q2: a) 2(d2 – 3d +3) = d + 1 b) 3(e + 1)2= 1 – e c) (g + 3)(2 – g) = g2
Answers • Q1: a) x = 2, x = - ½ b) x = 0, x= 9 c) x = -4, x=3 • Q2: a) (a+3)(a-3) b) e = -2, e = -1/3 c) d = -2, d = 1 ½
Final assessment • Solve by factorisation. 17x2 – 51x + 34 = 0
Hard Questions
11.7 – Quadratic Equations and Problem Solving A cliff diver is 64 feet above the surface of the water. The formula for calculating the height (h) of the diver after t seconds is: h  16t2 64. How long does it take for the diver to hit the surface of the water? 2 16t  64 16   2 t  4 16   2 t    2 t  0  0  0  t 2  0 t 2  0 t  2 t  2 seconds
11.7 – Quadratic Equations and Problem Solving The square of a number minus twice the number is 63. Find the number. x is the number. 2 x 2x  63  2 x  2x 63  0 63: Factors of 1, 63 3, 21 7, 9  x  7   9 x 0 x 7  0 x 9  0 x  7 x  9
11.7 – Quadratic Equations and Problem Solving The length of a rectangular garden is 5 feet more than its width. The area of the garden is 176 square feet. What are the length and the width of the garden? l wA The width is w. The length is w+5. w5w176   11 w  w11 0 w11 2 w 5w 176 2 w 5w176  0 w16  0 w 16 w11 l 115 l 16 feet feet Factors of 176: 1,176 2, 88 4, 44 8, 22 11,16   16 w  0 
11.7 – Quadratic Equations and Problem Solving Find two consecutive odd numbers whose product is 23 more than their sum? Consecutive odd numbers: x x 2. x  5x 5x 2 2 2 25 x x x    2 x  25 2 x  25  0 x5 50 x  50 x  52  3 5 2  7 5, 3 5, 7   2 x    2 x x   23 2 x  2x  2x  2x  25 2x 2 x  25  25 25 x5  0
11.7 – Quadratic Equations and Problem Solving The length of one leg of a right triangle is 7 meters less than the length of the other leg. The length of the hypotenuse is 13 meters. What are the lengths of the legs? ax a 12 Pythagorean Th.  2 2 2 x  x 7 13 x  5 meters  5 7bx 13 c  2 2 x  x 14x  49 169 2 2x 14x 120  0   2 2 x 7x 60  0 2 50x 12 0 x  x 12 b 127 meters 2 2 2 a b  c Factors of 60:1, 60 2, 30 3, 20 4,15 5,12   5 x   12 x  0 6,10

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Solving quadratic equations

  • 1. Equations and Quadratic functions Objective Solving quadratic equation by Factorisation
  • 2. Warm up • Solve the given equation.     y y y y y    3 6 1 2 6 3 11 5 3 5 11 3
  • 4. 11.6 – Solving Quadratic Equations by Factoring A quadratic equation is written in the Standard Form, ax2 bx  c  0 where a, b, and c are real numbers and . 0 a  Examples: 2 x 7x 12  0 xx 7  0 2 3x  4x 15 (standard form)
  • 5. 11.6 – Solving Quadratic Equations by Factoring Zero Factor Property: If a and b are real numbers and if , 0 ab  then or . 0 a  0 b  Examples: xx 7  0 x  0 x 7  0 x  0 x  7
  • 6. 11.6 – Solving Quadratic Equations by Factoring Zero Factor Property: If a and b are real numbers and if , 0 ab  then or . 0 a  0 b  Examples: x103x6  0 x 10  0 3x6  0 x 1010  010 3x 66  06 x 10 x  3 6 3 3 3x  6 x  2
  • 7. 11.6 – Solving Quadratic Equations by Factoring Solving Quadratic Equations: 1) Write the equation in standard form. 2) Factor the equation completely. 3) Set each factor equal to 0. 4) Solve each equation. 5) Check the solutions (in original equation).
  • 8. 11.6 – Solving Quadratic Equations by Factoring 2 x 3x 18 x2 3x 18  0 Factors of 18: 1, 18 2, 9 3, 6 x3 x 3  0 x 6  0 x  3 x  6     2 6 3 6 18 3618 18 18 18      2 3 3 3 18 99 18 18 18   6 x 0
  • 9. 11.6 – Solving Quadratic Equations by Factoring If the Zero Factor Property is not used, then the solutions will be incorrect 2 x 3x 18 xx3 18 x 18     2 18 3 18 18 32454 18 270 18      2 21 3 21 18 4416318 378 18 x318 x33183 x  21
  • 10. 11.6 – Solving Quadratic Equations by Factoring xx 4  5 x2  4x  5 2 x  4x 5  0 x 1x 5  0 x1 0 x 5  0 x  1 x  5
  • 11. 11.6 – Solving Quadratic Equations by Factoring x 33x 2  0 3x2  7x  6 30x 3 2 0 x  3 x  2 3 x  x3x 7  6 2 3 7 6 0 x x    32x  Factors of 3: 1, 3 Factors of 6: 1, 6 2, 3
  • 12. 11.6 – Solving Quadratic Equations by Factoring 9x2  24x  16 2 9x 24x 16  0 9 and 16 are perfect squares 3x 43x 4  0 3x4  0 3x  4 4 3 x 
  • 13. 11.6 – Solving Quadratic Equations by Factoring 2x3 18x  0 2x 2x   2 9 x  0    3 x    3 x  0  x 3  0 x3 0 2x  0 x  0 x  3 x  3
  • 14. 11.6 – Solving Quadratic Equations by Factoring x 33x2 20x 7  0 3: Factors of 1, 37 : Factors of 1, 7 x3 30 x    7 x  0   3 1 x  70x 3 1 0 x  x  3 x  7 3x  1 1 3 x  
  • 15. Continuous assessment • Q1: Solve these equations by factorisation. a) (x-2)(2x+1) = 0 b) 3x2 – 27x = 0 c) 2x2 – 7x + 6 = 0 Q2: a) 2(d2 – 3d +3) = d + 1 b) 3(e + 1)2= 1 – e c) (g + 3)(2 – g) = g2
  • 16. Answers • Q1: a) x = 2, x = - ½ b) x = 0, x= 9 c) x = -4, x=3 • Q2: a) (a+3)(a-3) b) e = -2, e = -1/3 c) d = -2, d = 1 ½
  • 17. Final assessment • Solve by factorisation. 17x2 – 51x + 34 = 0
  • 19. 11.7 – Quadratic Equations and Problem Solving A cliff diver is 64 feet above the surface of the water. The formula for calculating the height (h) of the diver after t seconds is: h  16t2 64. How long does it take for the diver to hit the surface of the water? 2 16t  64 16   2 t  4 16   2 t    2 t  0  0  0  t 2  0 t 2  0 t  2 t  2 seconds
  • 20. 11.7 – Quadratic Equations and Problem Solving The square of a number minus twice the number is 63. Find the number. x is the number. 2 x 2x  63  2 x  2x 63  0 63: Factors of 1, 63 3, 21 7, 9  x  7   9 x 0 x 7  0 x 9  0 x  7 x  9
  • 21. 11.7 – Quadratic Equations and Problem Solving The length of a rectangular garden is 5 feet more than its width. The area of the garden is 176 square feet. What are the length and the width of the garden? l wA The width is w. The length is w+5. w5w176   11 w  w11 0 w11 2 w 5w 176 2 w 5w176  0 w16  0 w 16 w11 l 115 l 16 feet feet Factors of 176: 1,176 2, 88 4, 44 8, 22 11,16   16 w  0 
  • 22. 11.7 – Quadratic Equations and Problem Solving Find two consecutive odd numbers whose product is 23 more than their sum? Consecutive odd numbers: x x 2. x  5x 5x 2 2 2 25 x x x    2 x  25 2 x  25  0 x5 50 x  50 x  52  3 5 2  7 5, 3 5, 7   2 x    2 x x   23 2 x  2x  2x  2x  25 2x 2 x  25  25 25 x5  0
  • 23. 11.7 – Quadratic Equations and Problem Solving The length of one leg of a right triangle is 7 meters less than the length of the other leg. The length of the hypotenuse is 13 meters. What are the lengths of the legs? ax a 12 Pythagorean Th.  2 2 2 x  x 7 13 x  5 meters  5 7bx 13 c  2 2 x  x 14x  49 169 2 2x 14x 120  0   2 2 x 7x 60  0 2 50x 12 0 x  x 12 b 127 meters 2 2 2 a b  c Factors of 60:1, 60 2, 30 3, 20 4,15 5,12   5 x   12 x  0 6,10