Algebra Demystified 2nd Ed

416 algebra De mystif ieD

The object will reach a height of 55 feet at about 0.35 seconds (on its

way up) and again at about 0.90 seconds (on its way down).

4. In the formula h = −16t^2 + v 0 t + h 0 , v 0 = 80 and h 0 = 36.
   h = −16t^2 + 80t + 36

We want to find t when h = 90.

90 16 80 36

0168054

1

2

0

2

2

()

= –+ +

= –+ –

• =

tt

tt

• –+ –

=  –  +  = 

1

2

16 80 54

0840 27 0

2

2

( )

tt

tt

t () ()()( )

()

= ––^40 ±  ––^40 4827 = ^ ±  –

28

(^2401600)
=  ± 
≈  ±  ≈ 
864
16
40 736
16
40 27 13
16
080420

 . ., .

The object will reach a height of 90 feet after about 0.80 seconds (on its

way up) and again at about 4.20 seconds (on its way down).

Problems in Geometry

To solve word problems involving geometric shapes, we begin with the formula

or formulas referred to in the problem. For example, after reading “The perim-

eter of a rectangular room.. .” we write P = 2 L + 2W. We then read the problem

and substitute the numbers given in the problem for the variables. For example,

after reading “The perimeter of the room is 50 feet.. .” we replace P with 50.

The formula then becomes and 50 = 2L + 2W. In the statement, “Its width is

two-thirds its length,” we write WL=^2

3

and the equation 50 = 2L + 2W

becomes 50 222

3

=+LL().

The formulas we need in this section are listed below.