- 1.
- Using Divide on the Polynomials
submenu,

=
x - 7 +
Thus, the remainder is 5k + 28. Applying Solve + Exact to 5k + 28 = 9
gives the solution
k = - 
.BITMAPSETProbSolvHint0.2006in0.243in0inq1
- 2.
- Defining functions
f (x) = x3 + x ln x and
g(x) = x + ex and evaluating gives
f (g(x)) |
= |
x + ex + x + ex ln x + ex |
|
g(f (x)) |
= |
x3 + x ln x + ex3+x ln x |
|
f (x)g(x) |
= |
x3 + x ln x x + ex |
|
f (x) + g(x) |
= |
x3 + x ln x + x + ex |
|
BITMAPSETProbSolvHint0.2006in0.243in0inq2
- 3.
- For any two distinct points
(x1, y1) and
(x2, y2) in the plane, there is a unique line ax + by + c = 0 through
these two points. Substituting these points in the equation for the line
gives the two equations
ax1 + by1 + c = 0 and
ax2 + by2 + c = 0. Apply
Exact from the Solve submenu to this system
ax1 + by1 + c |
= |
0 |
|
ax2 + by2 + c |
= |
0 |
|
of linear equations, solving for the variables a, b.
Solution is:
b = -
c
,
a =
c
Consequently, the equation for the line is
or, clearing fractions and collecting coefficients by factoring in place,
y1 -
y2
x -
x1 -
x2
y +
x1y2 -
y1x2
= 0
BITMAPSETProbSolvHint0.2006in0.243in0inq3
- 4.
- For the points
2, 5
,
3, - 7
, the system of equations is
2a + 5b + c |
= |
0 |
|
3a - 7b + c |
= |
0 |
|
Apply Exact from the Solve submenu to get
Solution is:
b = -
c,
a = -
c
Consequently, the equation for the line is
or, clearing fractions and simplifying,
-12x + y + 29 = 0
BITMAPSETProbSolvHint0.2006in0.243in0inq4
- 5.
- Since the point
0, 0
lies on the line,
you do not get a unique solution to the system of equations for the pair a, b. Thus, choosing Exact from the Solve submenu and
specifying a, b for the variables gives no response. However, specifying a, c for Variable(s) to Solve for gives the solution
a = - 2
b,
c = 0
Thus, the equation for the line is
-2bx + by = 0
or, dividing by b and applying Simplify,
Note: An interesting method for finding the equation of a line
through two specified points using determinants is described in Chapter 5,
Matrix Algebra.BITMAPSETProbSolvHint0.2006in0.243in0inq5
- 6.
- The slope-intercept form of the equation for a
line is y = mx + b, where m is the slope and b the y-intercept. If a
line is given as a linear equation in the form
sx + ty = c
you can find the slope by solving the equation for y. Apply Expand
to the solution
y = -
to get
y = -
sx +
c
revealing the slope to be
-
.BITMAPSETProbSolvHint0.2006in0.243in0inq6
- 7.
- To find the center and semi-axes of the ellipse
x2 +4y2 + 96x - 8y + 84 = 0,
You can read the answer from this form of the equation: The center of the
ellipse is (- 3, 1), and the semi-axes are
= 2 and
= 4.BITMAPSETProbSolvHint0.2006in0.243in0inq7
- 8.
- Apply Factor to several differences.
-
x2 - y2 =
x - y
x + y
-
x3 - y3 =
x - y
x2 + xy + y2
-
x4 - y4 =
x - y
x + y
x2 + y2
-
x5 - y5 =
x - y
x4 + x3y + x2y2 + xy3 + y4
-
x6 - y6 =
x - y
x + y
x2 + xy + y2
x2 - xy + y2
-
x7 - y7 =
x - y
x6 + x5y + x4y2 + x3y3 + x2y4 + xy5 + y6
After looking at only these few examples, you might find it reasonable to
conjecture that, for n odd,
We leave the general conjecture for you. Experiment.BITMAPSETProbSolvHint0.2006in0.243in0inq8
- 9.
- Using the clue from the example that the system will
factor over roots that appear as coefficients, apply Factor to the
product

x2 - 3
to get

x2 - 3
= 
x - 

x + 
. Now you can divide out the extraneous factor to get
For the polynomial
x3 +3x2 - 5x + 1, apply Roots from the
Polynomials submenu to find the roots:
1, -2 +
, -2 - 
. You can multiply by
to factor this
polynomial:

x3 +3x2 - 5x + 1
= 
x - 1
x + 2 + 

x + 2 - 
. Then,
dividing out the extraneous factor of
, you have
x3 +3
x2 -5
x + 1 =
x - 1

x + 2 +


x + 2 -

BITMAPSETProbSolvHint0.2006in0.243in0inq9
- 10.
- Analytic solution: The question asks when A = 2P. Solving the equation
2P = P
1 + r
for t gives the
solution
t =
Evaluate Numerically gives the number of years for each of the four
different interest rates as follows.
Experimental solution: For the interest rate 3%, you might start
with the following data:
Then, look at the years between 20 and 25:
These results indicate that it takes between 23 and 24 years to double an
investment at 3% annual interest. Take similar steps for the other interest
rates.BITMAPSETProbSolvHint0.2006in0.243in0inq10
- 11.
- With the insertion point in the array, from the
Solve submenu, choose Exact. You receive the response
Solution is : y = 1, x = 1 , y = ρ, x = - 3ρ3 + 4 |
where ρ is a root of 18Z5 +18Z4 +18Z3 -30Z2 - 30Z - 31 |
Leave your insertion point in the polynomial
18Z5 +18Z4 +18Z3 -30Z2 - 30Z - 31
and from the Polynomials submenu, choose Roots. You get
the following solution.
roots :
Define the function
x(t) = - 3t3 + 4 with New Definition from the
Define submenu. Select the vector of roots, and click the
parentheses icon. Type an x at the left of the vector, leave the insertion
point in the expression, and apply Evaluate to get the following:
Leave the insertion point in the matrix on the right and click Expand.
To display this result, you can concatenate the two vectors: Place them side
by side and from the Matrices submenu, choose Concatenate.
Then, select the (two-column) matrix, choose Edit + Insert Row(s)
to add a new first row, and label the columns with x and y as follows:
BITMAPSETProbSolvHint0.2006in0.243in0inq11
- 12.
- First, solve the system of equations for x and y
with Solve + Exact to get the solution
x = -
y =
(Note that the condition ad≠bc is necessary for this system of
equations to have a solution.) Now find the quadratic equation in u and v. To replace x and y with their equivalent expressions in terms of u
and v, edit the polynomial; or define the variables x and y and apply
Evaluate; or select the polynomial and from the Edit menu,
choose Replace and enter the appropriate expressions in
mathematics. These replacements get the following polynomial:
A -  + B -     + C  6pt |
+ D -  + E  + F = 0 |
To simplify this equation and get it into standard quadratic form in terms
of the variables u and v, first factor the left side of the equation to
find a common denominator. (This procedure gives a fraction too large to fit
well on the page
-
but you can still work with it on your screen.) Multiply by the denominator
ad - cb
to get the following quadratic equation in powers
of u and v:
Av2b2 -2Avbdu + Ad2u2 + Bvbcu - Bv2ba - Bdu2c
+ Bduav + Cc2u2 -2Ccuav + Ca2v2 + Dvb2c
- Dvbad - Dducb + Dd2ua + Ec2ub - Ecuad
- Eavcb + Ea2vd + Fc2b2 -2Fcbad + Fa2d2 = 0 |
Now, to identify the sign of the discriminant for this quadratic
equation, you need to collect coefficients of u and v. To collect them,
select and then click and drag terms containing a factor of u2 to the
beginning of the expression. Then, select all these terms, and while holding
down the CTRL key, choose Factor. Then, follow this same
procedure for uv, v2, u, and v in turn. You can get this equation
into the following form with these steps:
d2A - Bcd + c2C u2 + -2bdA + aBd + bBc - 2acC uv
+ - abB + Ab2 + a2C v2 + ad - cb Dd - cE u
+ ad - cb Ea - Db v + F - cb + ad = 0 |
To find the discriminant, replace
,
,
in the expression
=
-4
with these coefficients:
 |
= |
d2A - Bcd + c2C |
|
 |
= |
-2bdA + aBd + bBc - 2acC |
|
 |
= |
- abB + Ab2 + a2C |
|
By clicking and dragging, move terms having a factor of B2 adjacent to
one another. Repeat for terms having a factor of AC. Then, select all
terms having factors of B2 and, holding down the CTRL key,
factor in place. Repeat for all terms having factors of AC. Then, factor
the result. These successive steps are demonstrated in the following
display:
 |
= |
-4  |
|
|
= |
-2bdA + aBd + bBc - 2acC |
|
|
|
- 4 d2A - Bcd + c2C - abB + Ab2 + a2C |
|
|
= |
8bdAacC + a2B2d2 -2aB2dbc + b2B2c2 -4d2Aa2C - 4c2CAb2 |
|
|
= |
a2B2d2 -2aB2dbc + b2B2c2 +8bdAacC - 4d2Aa2C - 4c2CAb2 |
|
|
= |
B2 ad - cb -4AC ad - cb |
|
|
= |
B2 - 4AC ad - cb |
|
|
= |
 ad - cb |
|
It is easy to see from the equation
that
and
have the same sign
when ad≠bc.BITMAPSETProbSolvHint0.2006in0.243in0inq12