According
to Newtonian
mechanics,
this problem may be
expressed
as a second degree polynomial, where T is time in seconds.
Keys:
f(T)
=
(-9.81
+•
2)7*
+ 20T + 2
Display:
Description:
|XEQ|Q
2 0
Revalue
B?value
Starts the quadratic
equation program.
9.81
I+/-II
ENTER
I
IR/SI
Stores
(-9.81/2)
in A.
20
|R/S|
C?value
Stores
20
in
B.
2|R/S|
X=4.1751
Stores
2
in
C
and
cal
culates X (which in this
case
is
also
known
as
T).
|
R/S
|
X=-0.0977
Calculates
the
other
root.
Note that since a negative time has no meaning in the context of this
problem,
the
first
result,
4.1751
seconds,
is the
meaningful
answer.
.Example 4.
Find
the
roots
of the
following
second-degree
polyno
mial using the program as it is listed. Then change the sense of the
comparison at line Q12 so that the second root is computed first and
then the
results
are
compared.
Remember
to
restore
the
original
line
or clear the program when you
finish
this
example.
Keys:
XEQl Q
1[M]
3(T|6[r7s]
1
iR/sl
x2
+ (3 x
106)*
+1
= 0
Display:
Description:
lvalue
Starts program.
B?value
Stores
1
in
A.
C?value
Stores
3 x 106
in
B.
X=-3,008,000,00
Stores
1
in
C
and
cal
culates
the
first
root.
196
12:
Mathematics
Programs
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