Solving Quadratic Equations
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Transcript of Solving Quadratic Equations
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SOLVING QUADRATIC EQUATIONS
By Factoring(APPLICATIONS)Long Test 3 (20%) – Dec 12 (Mon)Summative Test (20%) – Dec 14 (Wed)
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Coverage of Long Test 3
•Solving Quadratic Equations by factoring
•Applications of factoring (word problems)
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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.
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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.
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Problem 1Find the whole number such that four times the number subtracted from three times the square of the number makes 15. Equation:
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Solution: Equation:
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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
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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)
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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.
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Seatwork: (Notebooks)NSM P. 105Numbers 3-6.
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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.
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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.
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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.
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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
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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.
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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
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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.
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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
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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.
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12. Two positive numbers differs by 5 and the square of their sum is 97. Find the numbers.4, 9
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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
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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