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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