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Solving Quadratic Equations

Solving Quadratic Equations. By Factoring (APPLICATIONS) Long Test 3 (20%) – Dec 12 (Mon) Summative Test (20%) – Dec 14 ( Wed ). Coverage of Long Test 3. Solving Quadratic Equations by factoring Applications of factoring (word problems). How to solve quadratic equation by factoring:.

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Solving Quadratic Equations

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  1. Solving Quadratic Equations By Factoring (APPLICATIONS) Long Test 3 (20%) – Dec 12 (Mon) Summative Test (20%) – Dec 14 (Wed)

  2. Coverage of Long Test 3 • Solving Quadratic Equations by factoring • Applications of factoring (word problems)

  3. How to solve quadratic equation by factoring: • Write the equation in standard form. • Factor the polynomial if possible. • Apply the zero product property by setting each factor equal to zero. • Solve for the variable.

  4. Standard Form Quadratic Equation Quadratic equations can be written in the form ax2 + bx + c = 0 where a, b, and c are real numbers with a  0. Standard form for a quadratic equation is in descending order equal to zero.

  5. Problem 1 Find the whole number such that four times the number subtracted from three times the square of the number makes 15. Equation:

  6. Solution: Equation:

  7. Problem 2 Find the whole number such that twice its square added to itself makes 10. 2x2 + x = 10 2x2 + x – 10 = 0 (2x + 5) (x – 2) = 0 x = -5 or x = 2 - answer 2

  8. Problem 3 p. 103 Find two consecutive positive odd numbers such that the sum of their squares is equal to 130. x = 1st odd # x + 2 = 2nd odd # x2 + (x + 2)2 = 130 (equation)

  9. Problem 4 , p. 104 The perimeter of a rectangle is 20 cm and its area is 24 cm2. Calculate the length and width of the rectangle.

  10. Seatwork: (Notebooks) NSM P. 105 Numbers 3-6.

  11. 7. A rectangular field, 70 m long and 50 m wide, has a path of uniform width around it. If the area of the path is 896 m2, find the width of the path.

  12. Solution: (#7) x Area of path = 896 m2 Area of inside = 3500 – 896 = 2604 m2 x 50 m x Area of field & path = 70 m x 50 m = 3500 m2 x 70 m (70 – 2x) (50 – 2x) = 2604 3500 – 240x + 4x2 - 2604 = 0 4x2 – 240x + 896 = 0 x2 – 60x + 224 = 0 ( x – 56) (x – 4) = 0 x = 56 or x = 4 The more sensible answer is 4 cm.

  13. 8. The base and height of a triangle are (x + 3) cm and (2x – 5) cm respectively, If the area of the triangle is 20 cm2, find x.

  14. Solution # 8. 2x - 5 A = 1bh 2 20 = 1 (x + 3)(2x – 5) 2 40 = (x + 3) (2x – 5) 40 = 2x2 + x – 15 0 = 2x2 + x - 55 0 = (2x + 11) (x – 5) 2x = -11 x = 5 2 x + 3

  15. 9. The difference between two numbers is 3. If the square of the smaller number is equal to 4 times the larger number, find the numbers.

  16. Solution # 9:x = larger numberx – 3 = smaller number (x – 3)2 = 4x x2 - 6x + 9 – 4x = 0 x2 - 10x + 9 = 0 (x – 9) (x – 1) = 0 x = 9 or x = 1 x – 3 = 6 x – 3 = -2

  17. 10. The length of a rectangle is 5 cm longer than its width and its area is 66 cm2. Find the perimeter of the rectangle.

  18. Solution # 10. A = 66 cm2 w w + 5 L = 11 cm w = 6 cm P = 2l + 2w P = 2(11) + 2(6) P = 22 + 12 P = 34 cm w(w + 5) = 66 w2 + 5w – 66 = 0 (w + 11) (w – 6) = 0 w = -11 or w = 6

  19. 11. Two positive numbers differ by 7 and the sum of their squares is 169. Find the numbers. x2 + (x – 7)2 = 169x2 + x2 – 14x + 49 – 169 = 0 2x2 - 14x – 120 = 0 x2 - 7x - 60 = 0 (x – 12) (x + 5) = 0x = 12 or x = -5Answer: The numbers are 12 and 5.

  20. 12. Two positive numbers differs by 5 and the square of their sum is 97. Find the numbers.4, 9

  21. 13. A piece of wire 44 cm long is cut into two parts and each part is bent to form a square. If the total area of the two squares is 65 cm2, find the perimeter of the two squares.16 cm, 28 cm

  22. 14. A particle is projected from ground level so that its height above the ground after t seconds is given by 20t – 5t2 m. After how many seconds is it 15 m above the ground? Can you explain briefly why there are two possible answers?1 or 3

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