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:

1.Write the equation in standard form.2.Factor the polynomial if possible.3.Apply the zero product property by setting each factor equal to zero.

4.Solve for the variable.

Standard Form Quadratic Equation

Quadratic equations can be written in the form

ax2 + bx + c = 0where a, b, and c are real numbers with a 0.

Standard form for a quadratic equation is in descending order equal to zero.

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

Solution: Equation:

Problem 2Find the whole number such that twice its square added to itself makes 10.2x2 + x = 102x2 + x – 10 = 0(2x + 5) (x – 2) = 0 x = -5 or x = 2 - answer 2

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)

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.

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

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.

Solution: (#7)x

xx

x

Area of path = 896 m2

Area of field & path = 70 m x 50 m = 3500 m2

Area of inside = 3500 – 896= 2604 m2

(70 – 2x) (50 – 2x) = 2604

50 m

70 m

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.

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.

Solution # 8.

2x - 5

x + 3A = 1 bh 220 = 1 (x + 3)(2x – 5) 240 = (x + 3) (2x – 5)40 = 2x2 + x – 15 0 = 2x2 + x - 55 0 = (2x + 11) (x – 5) 2x = -11 x = 5 2

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.

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

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.

Solution # 10.

w(w + 5) = 66w2 + 5w – 66 = 0(w + 11) (w – 6) = 0w = -11 or w = 6

L = 11 cm w = 6 cm

P = 2l + 2wP = 2(11) + 2(6)P = 22 + 12P = 34 cm

A = 66 cm2

w + 5

w

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) = 0 x = 12 or x = -5Answer: The numbers are 12 and 5.

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

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

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