Section 1.3

Quadratic Equations

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Quadratic Equations

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OBJECTIVE 1

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Solve the equation:

(a)

(b)

2 3 4 0x x

2 9 0x x

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Solve the equation: 2 4 4 0x x

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OBJECTIVE 2

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The Square Root Method

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Solve each equation.

(a) (b) 2 7x 23 9x

2 7

7 7

x

x or x

2( 3) 9

3 3 3 30 6

x

x or xx or x

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OBJECTIVE 3

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3. Finding the vertex of a Quadratic Function

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Example 4For t in seconds, the height of a object in feet is given by the formula y = f(t) = −16t2 + 32t + 16. Using algebra, find the maximum height reached by the object and the time that height is reached.

2 2

32 12 2( 16)

(1) 16 32 16 16(1) 32(1) 16 32

bt coordinate of vertexa

f t t feet

second

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2 2 2

80 202 2( 2)

(20) 2 80 2(20) 80(20) 800

80 2 80 2(20) 40

bw coordinate of vertexa

Area f w w meters

l w meters

Area = lw

OBJECTIVE 4

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13

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Find the real solutions, if any, of the equation:

(a)

(b)

(c) 22 5 3 0m m

2 5 6 0x x

2 4 2 0x x

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Find the real solutions, if any, of the equation: 22 3 2x x

2

2

2 2 3 02, 2, 3

( 2) ( 2) 4(2)(3) 2 4 24 2 202(2) 4 4

x xa b c

NO REAL SOLUTION

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OBJECTIVE 5

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2

2

2

144 9( 36 324)

36 324 16

36 308 014 22

Volume x x

x x

x xx or x

Solution is only x = 22

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• The area of the opening of a rectangular window is to be 306 square centimeters. If the length exceeds the width by 1 centimeter, what are the dimensions? (Page 122 #90)

• An object is propelled vertically upward with an initial velocity of 20 meters per second. The distance s of the object from the ground is s = -5t2 + 20t. (Page 123 #100)

(a) When will the object be 15 meters above the ground?

(b) When will it strike the ground?

(c) When will object reach a height of 100 meters?

Area = Length X width 306 = (w + 1)w

15 = -5t2 + 20t

0 = -5t2 + 20t

100 = -5t2 + 20t

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