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E XAMPLE 4 TAKS PRACTICE: Multiple Choice
✓ GUIDED PRACTICE for Examples 2, 3, and 4
Simplify the expression.
9. Î }
6
}
5
13.
26
} 7 2 Ï }
5
268 Chapter 4 Quadratic Functions and Factoring
What are the solutions of the equation 1 } 7 ( z 2 2) 2 5 6?
A 244, 40 B 22 7Ï }
6 , 2 1 7Ï }
6
C 22Ï }
42 , 2 1 Ï }
42 D 22 Ï} 42
} ,21
7 Ï} 42
}
7
Solution
1
} 7 ( z 2 2) 2 5 6 Write original equation.
(z 2 2) 2 5 42 Multiply each side by 7.
z 2 2 56Ï }
42 Take square roots of each side.
Solve the equation.
z 5 2 6 Ï }
42 Add 2 to each side.
The solutions are 2 1 Ï }
42 and 2 2 Ï }
42 .
c The correct answer is C. A B C D
10. Î }
9
} 8
14.
2
}
4 1 Ï }
11
11. Î }
17
}
12
15.
21
} 9 1 Ï }
7
12. Î }
19
}
21
16.
4
}
8 2 Ï }
3
17. 5x 2 5 80 18. z 2 2 7 5 29 19. 3(x 2 2) 2 5 40
MODELING DROPPED OBJECTS When an
object is dropped, its height h (in feet)
above the ground after t seconds can be
modeled by the function
h 5216t 2 1 h 0
where h 0 is the object’s initial height
(in feet). The graph of h 5216t 2 1 200,
representing the height of an object
dropped from an initial height of 200 feet,
is shown at the right.
The model h 5216t 2 1 h 0 assumes that
the force of air resistance on the object is
negligible. Also, this model works only on
Earth. For planets with stronger or weaker
gravity, different models are used (see
Exercise 39).