Solutions

1.
Using Divide on the Polynomials submenu,

$\displaystyle {\frac{{x^{2}-3x+5k}}{{x+4}}}$ =  x - 7 + $\displaystyle {\frac{{5k+28}}{{x+4}}}$

Thus, the remainder is 5k + 28. Applying Solve + Exact to 5k + 28 = 9 gives the solution $\left\{\vphantom{ k=-\frac{19}{5}}\right.$k = - ${\frac{{19}}{{5}}}$$\left.\vphantom{ k=-\frac{19}{5}}\right\}$.BITMAPSETProbSolvHint0.2006in0.243in0inq1

2.
Defining functions f (x) = x3 + x ln x and g(x) = x + ex and evaluating gives
f (g(x)) = $\displaystyle \left(\vphantom{ x+e^{x}}\right.$x + ex$\displaystyle \left.\vphantom{ x+e^{x}}\right)^{{3}}_{}$ + $\displaystyle \left(\vphantom{ x+e^{x}}\right.$x + ex$\displaystyle \left.\vphantom{ x+e^{x}}\right)$ln$\displaystyle \left(\vphantom{ x+e^{x}}\right.$x + ex$\displaystyle \left.\vphantom{ x+e^{x}}\right)$  
g(f (x)) = x3 + x ln x + ex3+x ln x  
f (x)g(x) = $\displaystyle \left(\vphantom{ x^{3}+x\ln x}\right.$x3 + x ln x$\displaystyle \left.\vphantom{ x^{3}+x\ln x}\right)$$\displaystyle \left(\vphantom{ x+e^{x}}\right.$x + ex$\displaystyle \left.\vphantom{ x+e^{x}}\right)$  
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: $\displaystyle \left\{\vphantom{ b=-c\dfrac{x_{1}-x_{2}}{-y_{1}x_{2}+x_{1}y_{2}}%
,a=c\dfrac{y_{1}-y_{2}}{-y_{1}x_{2}+x_{1}y_{2}}}\right.$b = - c$\displaystyle {\dfrac{{x_{1}-x_{2}}}{{-y_{1}x_{2}+x_{1}y_{2}}%
}}$, a = c$\displaystyle {\dfrac{{y_{1}-y_{2}}}{{-y_{1}x_{2}+x_{1}y_{2}}}}$$\displaystyle \left.\vphantom{ b=-c\dfrac{x_{1}-x_{2}}{-y_{1}x_{2}+x_{1}y_{2}}%
,a=c\dfrac{y_{1}-y_{2}}{-y_{1}x_{2}+x_{1}y_{2}}}\right\}$

Consequently, the equation for the line is

c$\displaystyle \left(\vphantom{ \frac{y_{1}-y_{2}}{-y_{1}x_{2}+x_{1}y_{2}}}\right.$$\displaystyle {\frac{{y_{1}-y_{2}}}{{-y_{1}x_{2}+x_{1}y_{2}}}}$$\displaystyle \left.\vphantom{ \frac{y_{1}-y_{2}}{-y_{1}x_{2}+x_{1}y_{2}}}\right)$x - c$\displaystyle \left(\vphantom{ \frac{%
x_{1}-x_{2}}{-y_{1}x_{2}+x_{1}y_{2}}}\right.$$\displaystyle {\frac{{%
x_{1}-x_{2}}}{{-y_{1}x_{2}+x_{1}y_{2}}}}$$\displaystyle \left.\vphantom{ \frac{%
x_{1}-x_{2}}{-y_{1}x_{2}+x_{1}y_{2}}}\right)$y + c = 0

or, clearing fractions and collecting coefficients by factoring in place,

$\displaystyle \left(\vphantom{ y_{1}-y_{2}}\right.$y1 - y2$\displaystyle \left.\vphantom{ y_{1}-y_{2}}\right)$x - $\displaystyle \left(\vphantom{ x_{1}-x_{2}}\right.$x1 - x2$\displaystyle \left.\vphantom{ x_{1}-x_{2}}\right)$y + $\displaystyle \left(\vphantom{
x_{1}y_{2}-y_{1}x_{2}}\right.$x1y2 - y1x2$\displaystyle \left.\vphantom{
x_{1}y_{2}-y_{1}x_{2}}\right)$ = 0

BITMAPSETProbSolvHint0.2006in0.243in0inq3

4.
For the points $\left(\vphantom{ 2,5}\right.$2, 5$\left.\vphantom{ 2,5}\right)$, $\left(\vphantom{ 3,-7}\right.$3, - 7$\left.\vphantom{ 3,-7}\right)$, the system of equations is
2a + 5b + c = 0  
3a - 7b + c = 0  

Apply Exact from the Solve submenu to get

Solution is: $\displaystyle \left\{\vphantom{ b=-\frac{1}{29}c,a=-\frac{12}{29}c}\right.$b = - $\displaystyle {\frac{{1}}{{29}}}$c, a = - $\displaystyle {\frac{{12}}{{29}}}$c$\displaystyle \left.\vphantom{ b=-\frac{1}{29}c,a=-\frac{12}{29}c}\right\}$

Consequently, the equation for the line is

- $\displaystyle {\frac{{12}}{{29}}}$cx - $\displaystyle \left(\vphantom{ -\frac{1}{29}}\right.$ - $\displaystyle {\frac{{1}}{{29}}}$$\displaystyle \left.\vphantom{ -\frac{1}{29}}\right)$cy + c = 0

or, clearing fractions and simplifying,

-12x + y + 29 = 0

BITMAPSETProbSolvHint0.2006in0.243in0inq4

5.
Since the point $\left(\vphantom{ 0,0}\right.$0, 0$\left.\vphantom{ 0,0}\right)$ 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

$\displaystyle \left\{\vphantom{ a=-2b,c=0}\right.$a = - 2b, c = 0$\displaystyle \left.\vphantom{ a=-2b,c=0}\right\}$

Thus, the equation for the line is

-2bx + by = 0

or, dividing by b and applying Simplify,

$\displaystyle \left(\vphantom{ -2bx+by}\right.$ -2bx + by$\displaystyle \left.\vphantom{ -2bx+by}\right)$$\displaystyle {\frac{{1}}{{b}}}$ = - 2x + y = 0

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 = - $\displaystyle {\frac{{sx-c}}{{t}}}$

to get

y = - $\displaystyle {\frac{{1}}{{t}}}$sx + $\displaystyle {\frac{{1}}{{t}}}$c

revealing the slope to be - ${\dfrac{{s}}{{t}}}$.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 $\sqrt{{4}}$ =  2 and $\sqrt{{16}}$ =  4.BITMAPSETProbSolvHint0.2006in0.243in0inq7

8.
Apply Factor to several differences.

x2 - y2 =  $\left(\vphantom{ x-y}\right.$x - y$\left.\vphantom{ x-y}\right)$$\left(\vphantom{ x+y}\right.$x + y$\left.\vphantom{ x+y}\right)$

x3 - y3 =  $\left(\vphantom{ x-y}\right.$x - y$\left.\vphantom{ x-y}\right)$$\left(\vphantom{ x^{2}+xy+y^{2}}\right.$x2 + xy + y2$\left.\vphantom{ x^{2}+xy+y^{2}}\right)$

x4 - y4 =  $\left(\vphantom{ x-y}\right.$x - y$\left.\vphantom{ x-y}\right)$$\left(\vphantom{ x+y}\right.$x + y$\left.\vphantom{ x+y}\right)$$\left(\vphantom{
x^{2}+y^{2}}\right.$x2 + y2$\left.\vphantom{
x^{2}+y^{2}}\right)$

x5 - y5 =  $\left(\vphantom{ x-y}\right.$x - y$\left.\vphantom{ x-y}\right)$$\left(\vphantom{
x^{4}+x^{3}y+x^{2}y^{2}+xy^{3}+y^{4}}\right.$x4 + x3y + x2y2 + xy3 + y4$\left.\vphantom{
x^{4}+x^{3}y+x^{2}y^{2}+xy^{3}+y^{4}}\right)$

x6 - y6 =  $\left(\vphantom{ x-y}\right.$x - y$\left.\vphantom{ x-y}\right)$$\left(\vphantom{ x+y}\right.$x + y$\left.\vphantom{ x+y}\right)$$\left(\vphantom{ x^{2}+xy+y^{2}}\right.$x2 + xy + y2$\left.\vphantom{ x^{2}+xy+y^{2}}\right)$$\left(\vphantom{ x^{2}-xy+y^{2}}\right.$x2 - xy + y2$\left.\vphantom{ x^{2}-xy+y^{2}}\right)$

x7 - y7 =  $\left(\vphantom{ x-y}\right.$x - y$\left.\vphantom{ x-y}\right)$$\left(\vphantom{
x^{6}+x^{5}y+x^{4}y^{2}+x^{3}y^{3}+x^{2}y^{4}+xy^{5}+y^{6}}\right.$x6 + x5y + x4y2 + x3y3 + x2y4 + xy5 + y6$\left.\vphantom{
x^{6}+x^{5}y+x^{4}y^{2}+x^{3}y^{3}+x^{2}y^{4}+xy^{5}+y^{6}}\right)$

After looking at only these few examples, you might find it reasonable to conjecture that, for n odd,

xn - yn =  $\displaystyle \left(\vphantom{ x-y}\right.$x - y$\displaystyle \left.\vphantom{ x-y}\right)$$\displaystyle \sum_{{k=0}}^{{n-1}}$xn-k-1yk

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 $\sqrt{{3}}$$\left(\vphantom{ x^{2}-3}\right.$x2 - 3$\left.\vphantom{ x^{2}-3}\right)$ to get $\sqrt{{3}}$$\left(\vphantom{ x^{2}-3}\right.$x2 - 3$\left.\vphantom{ x^{2}-3}\right)$ =  $\sqrt{{3}}$$\left(\vphantom{ x-\sqrt{3}}\right.$x - $\sqrt{{3}}$$\left.\vphantom{ x-\sqrt{3}}\right)$$\left(\vphantom{ x+\sqrt{3}}\right.$x + $\sqrt{{3}}$$\left.\vphantom{ x+\sqrt{3}}\right)$. Now you can divide out the extraneous factor to get

x2 -3 =  $\displaystyle \left(\vphantom{ x-\sqrt{3}}\right.$x - $\displaystyle \sqrt{{3}}$$\displaystyle \left.\vphantom{ x-\sqrt{3}}\right)$$\displaystyle \left(\vphantom{ x+\sqrt{3}}\right.$x + $\displaystyle \sqrt{{3}}$$\displaystyle \left.\vphantom{ x+\sqrt{3}}\right)$

For the polynomial x3 +3x2 - 5x + 1, apply Roots from the Polynomials submenu to find the roots: $\left\{\vphantom{ 1,-2+\sqrt{5},-2-%
\sqrt{5}}\right.$1, -2 + $\sqrt{{5}}$, -2 - $\sqrt{{5}}$$\left.\vphantom{ 1,-2+\sqrt{5},-2-%
\sqrt{5}}\right\}$. You can multiply by $\sqrt{{5}}$ to factor this polynomial: $\sqrt{{5}}$$\left(\vphantom{ x^{3}+3x^{2}-5x+1}\right.$x3 +3x2 - 5x + 1$\left.\vphantom{ x^{3}+3x^{2}-5x+1}\right)$  =  $\sqrt{{5}}$$\left(\vphantom{
x-1}\right.$x - 1$\left.\vphantom{
x-1}\right)$$\left(\vphantom{ x+2+\sqrt{5}}\right.$x + 2 + $\sqrt{{5}}$$\left.\vphantom{ x+2+\sqrt{5}}\right)$$\left(\vphantom{ x+2-\sqrt{5}}\right.$x + 2 - $\sqrt{{5}}$$\left.\vphantom{ x+2-\sqrt{5}}\right)$. Then, dividing out the extraneous factor of $\sqrt{{5}}$, you have

x3 +3x2 -5x + 1  =  $\displaystyle \left(\vphantom{ x-1}\right.$x - 1$\displaystyle \left.\vphantom{ x-1}\right)$$\displaystyle \left(\vphantom{ x+2+\sqrt{5}}\right.$x + 2 + $\displaystyle \sqrt{{5}}$$\displaystyle \left.\vphantom{ x+2+\sqrt{5}}\right)$$\displaystyle \left(\vphantom{
x+2-\sqrt{5}}\right.$x + 2 - $\displaystyle \sqrt{{5}}$$\displaystyle \left.\vphantom{
x+2-\sqrt{5}}\right)$

BITMAPSETProbSolvHint0.2006in0.243in0inq9

10.
Analytic solution: The question asks when A = 2P. Solving the equation 2P = P$\left(\vphantom{ 1+r}\right.$1 + r$\left.\vphantom{ 1+r}\right)^{{t}}_{}$ for t gives the solution

t = $\displaystyle {\frac{{\ln 2}}{{\ln \left( 1+r\right) }}}$

Evaluate Numerically gives the number of years for each of the four different interest rates as follows.

$\displaystyle \fbox{$
\begin{array}{c}
\text{For 3\%, \ }\dfrac{\ln 2}{\ln \l...
... }\dfrac{\ln 2}{\ln \left( 1+.10\right) }=7.27\text{ years}
\end{array}
$}
$

  

Experimental solution: For the interest rate 3%, you might start with the following data:

$\displaystyle \fbox{$
\begin{array}{cc}
10\text{ years:} & A=P\left( 1+.03\ri...
...
25\text{ years:} & A=P\left( 1+.03\right) ^{25}=\,2.09P
\end{array}
$}
$

Then, look at the years between 20 and 25: 

$\displaystyle \fbox{$
\begin{array}{cc}
21\text{ years:} & A=P\left( 1+.03\ri...
...
24\text{ years:} & A=P\left( 1+.03\right) ^{24}=\,2.03P
\end{array}
$}
$

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 : $\displaystyle \left\{\vphantom{ y=1,x=1}\right.$y = 1, x = 1$\displaystyle \left.\vphantom{ y=1,x=1}\right\}$$\displaystyle \left\{\vphantom{ y=\rho ,x=-3\rho
^{3}+4}\right.$y = ρ, x = - 3ρ3 + 4$\displaystyle \left.\vphantom{ y=\rho ,x=-3\rho
^{3}+4}\right\}$
where ρ is a root of $\displaystyle \left(\vphantom{
18Z^{5}+18Z^{4}+18Z^{3}-30Z^{2}-30Z-31}\right.$18Z5 +18Z4 +18Z3 -30Z2 - 30Z - 31$\displaystyle \left.\vphantom{
18Z^{5}+18Z^{4}+18Z^{3}-30Z^{2}-30Z-31}\right)$

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 : $\displaystyle \fbox{$
\begin{array}{c}
-.60663-.98268i \\
-.60663+.98268i \\
-.48801-.92069i \\
-.48801+.92069i \\
1.1893
\end{array}
$}
$

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:

x$\displaystyle \left(\vphantom{
\begin{array}{c}
-.60663-.98268i \\
-.6066...
...\\
-.48801-.92069i \\
-.48801+.92069i \\
1.1893
\end{array}
}\right.$$\displaystyle \begin{array}{c}
-.60663-.98268i \\
-.60663+.98268i \\
-.48801-.92069i \\
-.48801+.92069i \\
1.1893
\end{array}$$\displaystyle \left.\vphantom{
\begin{array}{c}
-.60663-.98268i \\
-.6066...
...\\
-.48801-.92069i \\
-.48801+.92069i \\
1.1893
\end{array}
}\right)$ =  $\displaystyle \left(\vphantom{
\begin{array}{c}
-3\left( -.60663-.98268i\rig...
...
-3\left( -.48801+.92069i\right) ^{3}+4 \\
-1.0466
\end{array}
}\right.$$\displaystyle \begin{array}{c}
-3\left( -.60663-.98268i\right) ^{3}+4 \\
-3...
... ^{3}+4 \\
-3\left( -.48801+.92069i\right) ^{3}+4 \\
-1.0466
\end{array}$$\displaystyle \left.\vphantom{
\begin{array}{c}
-3\left( -.60663-.98268i\rig...
...
-3\left( -.48801+.92069i\right) ^{3}+4 \\
-1.0466
\end{array}
}\right)$

Leave the insertion point in the matrix on the right and click Expand.

$\displaystyle \left(\vphantom{
\begin{array}{c}
-3\left( -.60663-.98268i\rig...
...
-3\left( -.48801+.92069i\right) ^{3}+4 \\
-1.0466
\end{array}
}\right.$$\displaystyle \begin{array}{c}
-3\left( -.60663-.98268i\right) ^{3}+4 \\
-3...
... ^{3}+4 \\
-3\left( -.48801+.92069i\right) ^{3}+4 \\
-1.0466
\end{array}$$\displaystyle \left.\vphantom{
\begin{array}{c}
-3\left( -.60663-.98268i\rig...
...
-3\left( -.48801+.92069i\right) ^{3}+4 \\
-1.0466
\end{array}
}\right)$ =  $\displaystyle \left(\vphantom{
\begin{array}{c}
-.60247+.40783i \\
-.6024...
... \\
.62562-.36793i \\
.62562+.36793i \\
-1.0466
\end{array}
}\right.$$\displaystyle \begin{array}{c}
-.60247+.40783i \\
-.60247-.40783i \\
.62562-.36793i \\
.62562+.36793i \\
-1.0466
\end{array}$$\displaystyle \left.\vphantom{
\begin{array}{c}
-.60247+.40783i \\
-.6024...
... \\
.62562-.36793i \\
.62562+.36793i \\
-1.0466
\end{array}
}\right)$

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:

$\displaystyle \fbox{$
\begin{array}{cc}
\mathbf{x=-3y}^{3}\mathbf{+4} & \math...
...
.62562+.36793i & -.48801+.92069i \\
-1.0466 & 1.1893
\end{array}
$}
$

BITMAPSETProbSolvHint0.2006in0.243in0inq11

12.
First, solve the system of equations for x and y with Solve + Exact to get the solution

x = - $\displaystyle {\dfrac{{bv-du}}{{ad-bc}}}$        y = $\displaystyle {\dfrac{{av-cu}}{{ad-bc}}}$

(Note that the condition adbc 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$\displaystyle \left(\vphantom{ -\dfrac{vb-du}{-cb+ad}}\right.$ - $\displaystyle {\dfrac{{vb-du}}{{-cb+ad}}}$$\displaystyle \left.\vphantom{ -\dfrac{vb-du}{-cb+ad}}\right)^{{2}}_{}$ + B$\displaystyle \left(\vphantom{ -\dfrac{vb-du}{-cb+ad}%
}\right.$ - $\displaystyle {\dfrac{{vb-du}}{{-cb+ad}%
}}$$\displaystyle \left.\vphantom{ -\dfrac{vb-du}{-cb+ad}%
}\right)$$\displaystyle \left(\vphantom{ \dfrac{-cu+av}{-cb+ad}}\right.$$\displaystyle {\dfrac{{-cu+av}}{{-cb+ad}}}$$\displaystyle \left.\vphantom{ \dfrac{-cu+av}{-cb+ad}}\right)$ + C$\displaystyle \left(\vphantom{ \dfrac{-cu+av}{-cb+ad}%
}\right.$$\displaystyle {\dfrac{{-cu+av}}{{-cb+ad}%
}}$$\displaystyle \left.\vphantom{ \dfrac{-cu+av}{-cb+ad}%
}\right)^{{2}}_{}$ 6pt
+ D$\displaystyle \left(\vphantom{ -\dfrac{vb-du}{-cb+ad}}\right.$ - $\displaystyle {\dfrac{{vb-du}}{{-cb+ad}}}$$\displaystyle \left.\vphantom{ -\dfrac{vb-du}{-cb+ad}}\right)$ + E$\displaystyle \left(\vphantom{ \dfrac{-cu+av}{-cb+ad}%
}\right.$$\displaystyle {\dfrac{{-cu+av}}{{-cb+ad}%
}}$$\displaystyle \left.\vphantom{ \dfrac{-cu+av}{-cb+ad}%
}\right)$ + 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

- $\displaystyle {\frac{{-Av^{2}b^{2}+2Avbdu-Ad^{2}u^{2}-\cdots -Fc^{2}b^{2}+2Fcbad-Fa^{2}d^{2}%
}}{{\left( ad-cb\right) ^{2}}}}$

but you can still work with it on your screen.) Multiply by the denominator $\left(\vphantom{ ad-cb}\right.$ad - cb$\left.\vphantom{ ad-cb}\right)^{{2}}_{}$ 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:

$\left(\vphantom{ d^{2}A-Bcd+c^{2}C}\right.$d2A - Bcd + c2C$\left.\vphantom{ d^{2}A-Bcd+c^{2}C}\right)$u2 + $\left(\vphantom{ -2bdA+aBd+bBc-2acC}\right.$ -2bdA + aBd + bBc - 2acC$\left.\vphantom{ -2bdA+aBd+bBc-2acC}\right)$uv          + $\left(\vphantom{ -abB+Ab^{2}+a^{2}C}\right.$ - abB + Ab2 + a2C$\left.\vphantom{ -abB+Ab^{2}+a^{2}C}\right)$v2 + $\left(\vphantom{ ad-cb}\right.$ad - cb$\left.\vphantom{ ad-cb}\right)$$\left(\vphantom{
Dd-cE}\right.$Dd - cE$\left.\vphantom{
Dd-cE}\right)$u          + $\left(\vphantom{ ad-cb}\right.$ad - cb$\left.\vphantom{ ad-cb}\right)$$\left(\vphantom{ Ea-Db}\right.$Ea - Db$\left.\vphantom{ Ea-Db}\right)$v +  F$\left(\vphantom{ -cb+ad}\right.$ - cb + ad$\left.\vphantom{ -cb+ad}\right)^{{2}}_{}$ = 0

To find the discriminant, replace $\widetilde{{A}}$,$\widetilde{{B}}$,$\widetilde{{C}}$ in the expression $\widetilde{{\bigtriangleup }}$ = $\widetilde{{B}}^{{2}}_{}$ -4$\widetilde{{%
A}}$$\widetilde{{C}}$ with these coefficients:
$\displaystyle \widetilde{{A}}$ = d2A - Bcd + c2C  
$\displaystyle \widetilde{{B}}$ = -2bdA + aBd + bBc - 2acC  
$\displaystyle \widetilde{{C}}$ = - 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:
$\displaystyle \widetilde{{\bigtriangleup }}$ = $\displaystyle \widetilde{{B}}^{{2}}_{}$ -4$\displaystyle \widetilde{{A}}$$\displaystyle \widetilde{{C}}$  
         = $\displaystyle \left(\vphantom{ -2bdA+aBd+bBc-2acC}\right.$ -2bdA + aBd + bBc - 2acC$\displaystyle \left.\vphantom{ -2bdA+aBd+bBc-2acC}\right)^{{2}}_{}$  
                     -  4$\displaystyle \left(\vphantom{ d^{2}A-Bcd+c^{2}C}\right.$d2A - Bcd + c2C$\displaystyle \left.\vphantom{ d^{2}A-Bcd+c^{2}C}\right)$$\displaystyle \left(\vphantom{
-abB+Ab^{2}+a^{2}C}\right.$ - abB + Ab2 + a2C$\displaystyle \left.\vphantom{
-abB+Ab^{2}+a^{2}C}\right)$  
         =  8bdAacC + a2B2d2 -2aB2dbc + b2B2c2 -4d2Aa2C - 4c2CAb2  
         = a2B2d2 -2aB2dbc + b2B2c2 +8bdAacC - 4d2Aa2C - 4c2CAb2  
         = B2$\displaystyle \left(\vphantom{ ad-cb}\right.$ad - cb$\displaystyle \left.\vphantom{ ad-cb}\right)^{{2}}_{}$ -4AC$\displaystyle \left(\vphantom{ ad-cb}\right.$ad - cb$\displaystyle \left.\vphantom{ ad-cb}\right)^{{2}}_{}$  
         = $\displaystyle \left(\vphantom{ B^{2}-4AC}\right.$B2 - 4AC$\displaystyle \left.\vphantom{ B^{2}-4AC}\right)$$\displaystyle \left(\vphantom{ ad-cb}\right.$ad - cb$\displaystyle \left.\vphantom{ ad-cb}\right)^{{2}}_{}$  
         = $\displaystyle \bigtriangleup$$\displaystyle \left(\vphantom{ ad-cb}\right.$ad - cb$\displaystyle \left.\vphantom{ ad-cb}\right)^{{2}}_{}$  

It is easy to see from the equation

$\displaystyle \widetilde{{\bigtriangleup }}$ = $\displaystyle \bigtriangleup$$\displaystyle \left(\vphantom{ ad-cb}\right.$ad - cb$\displaystyle \left.\vphantom{ ad-cb}\right)^{{2}}_{}$

that $\bigtriangleup$ and $\widetilde{{\bigtriangleup }}$ have the same sign when adbc.BITMAPSETProbSolvHint0.2006in0.243in0inq12



Subsections