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{SECT 0 {PARA 0 "" 0 "" {TEXT 263 46 "High School Modules > Algebra Gr
egory A. Moore" }}{PARA 3 "" 0 "" {TEXT -1 4 "    " }{TEXT 262 26 "Rat
ionalizing Denominators" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 
0 "" {TEXT -1 66 "How to remove terms with radicals from denominators \+
of expressions" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT 264 153 "[Directions : Execute the Code Resource section first. \+
Although there will be no output immediately, these definitions are us
ed later in this worksheet.]" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 
0 {PARA 4 "" 0 "" {TEXT -1 7 "0. Code" }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 9 "restart; " }}}}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 37 "1.
 Difference of Squares & Conjugates" }}{PARA 0 "" 0 "" {TEXT -1 157 "
\nYou probably call the factoring formula called a \"Difference of Squ
ares\" or \"Product of a sum and difference of the same terms\". Lets \+
re-examine that idea.\n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "(
a+b)*(a-b): % = expand(%);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
27 "(x+7)*(x-7): % = expand(%);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 33 "(5*x+14)*(5*x-14): % = expand(%);" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 43 "(331*x + 214)*(331*x - 214): % = expand(%);" }}}
{PARA 0 "" 0 "" {TEXT -1 304 "\nThere is a pattern here. Whenever you \+
multiply two binomial terms, which are identical except for the middle
 sign being opposite, we get this same result. Although these two term
s are not exactly the same, they form a pair, and our called \"conjuga
tes\". One is the conjugate of the other, and vice-versa." }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "3*x + 5; Conjugate = op(1,%) - op(2
,%);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "3*x - 5; Conjugate \+
= op(1,%) - op(2,%);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "70*
x  - 3*sqrt(19); Conjugate = op(1,%) - op(2,%);" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 47 "100 + sqrt(5)*x; Conjugate = op(1,%) - op(2,%)
;" }}}{PARA 0 "" 0 "" {TEXT -1 119 "\nWhen applied to terms with squar
e roots, conjugates have a wonderous affect  - the result in radical f
ree expressions!" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 59 "(sqrt(15
) + sqrt(11))*(sqrt(15) - sqrt(11)): % = expand(%);" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 63 "(sqrt(96) + 3*sqrt(19))*(sqrt(96) - 3*sqr
t(19)): % = expand(%);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "(
 11 + sqrt(37))*( 11 - sqrt(37)): % = expand(%);" }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 65 "(7*sqrt(30) + sqrt(210))*(7*sqrt(30) - sqrt(2
10)): % = expand(%);" }}}{PARA 0 "" 0 "" {TEXT -1 50 "\nThis always wo
rks. Thus it can be a useful trick." }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 61 "( sqrt(a) + sqrt(b) ) * ( sqrt(a) - sqrt(b) ): % = ex
pand(%);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 69 "( c*sqrt(a) + d
*sqrt(b) ) * ( c*sqrt(a) - d*sqrt(b) ): % = expand(%);" }}}}{SECT 0 
{PARA 4 "" 0 "" {TEXT -1 45 "2. Denominators with One Radical - Square
root" }}{PARA 0 "" 0 "" {TEXT -1 168 "\n Now lets look at fractional e
xpressions which squareroots in their denominators. These expressions \+
are not considered to be in good form and need to be simplified.\n\n[
" }{TEXT 261 4 "Note" }{TEXT -1 388 " : Maple is so inherently clever \+
that it automatically rationalizes these problems. Just by entering th
e expression and executing it, Maple will rationalize it without any e
xplicit direction from you. However, our focus here is to understand t
he process of rationalizing the denominator in particular. This needs \+
to be accompanied by board work to demonstrate the method in more deta
il.]\n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "1/sqrt(7);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "denominator := sqrt(7);" }}}
{PARA 0 "" 0 "" {TEXT -1 86 "We can ratonalize the denominator by simp
ly multiplying by the same square root again." }}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 26 "denominator * denominator;" }}}{PARA 0 "" 0 "" 
{TEXT -1 25 "\nHere is another example." }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 15 "1/(2*sqrt(11));" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 26 "denominator := 2*sqrt(11);" }}}{PARA 0 "" 0 "" {TEXT 
-1 103 "\nWe can use the same trick again ... multiplying by the denom
inator on top and bottom ...and it works. " }}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 27 "denominator * (2*sqrt(11));" }}}{PARA 0 "" 0 "" 
{TEXT -1 337 "However, it's a little bit of an overkill. The 2 in the \+
denominator was already a rational number, so there was no need to mul
tiply by it. The only \"offensive\" part of the denominator is the squ
are root of 11. So we multiply by that square root we keep the numbers
 smaller, and make simpler multiplication and less simplification late
r. " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "denominator * sqrt(11
);" }}}{PARA 0 "" 0 "" {TEXT -1 1 "\n" }}}{SECT 0 {PARA 4 "" 0 "" 
{TEXT -1 47 "3. Denominators with One Radical - Higher Roots" }}{PARA 
0 "" 0 "" {TEXT -1 2 "\n " }{TEXT 259 11 "Example 3.1" }{TEXT -1 36 " \+
: Let's look at problems like this " }{XPPEDIT 18 0 "1/(5^(1/3));" "6#
*&\"\"\"F$)\"\"&*&F$F$\"\"$!\"\"F)" }{TEXT -1 141 ". \n\nHere is the d
enominator. But if we try the same trick of multiplying by the same ro
ot again, we don't end up with a root-free expression." }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "denominator := 5^(1/3);" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "denominator * 5^(1/3);" }}}{PARA 0 
"" 0 "" {TEXT -1 139 "\nThe key to unlocking a cube root is a cube. If
 we have one five under the radical, how many more do we need to get a
 total of three fives?" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "de
nominator *  (5^2)^(1/3): % = simplify(%);" }}}{PARA 0 "" 0 "" {TEXT 
-1 94 "That's the trick! Multiply top and bottom by the cuberoot of (5
^2) - not by the cuberoot of 5!" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 12 "1/(5^(1/3));" }}}{PARA 0 "" 0 "" {TEXT -1 29 "\n\nHere is anot
her example :\n\n" }{TEXT 260 11 "Example 3.2" }{TEXT -1 34 " : Ration
alize the denominator of " }{XPPEDIT 18 0 "(3/4)^(1/3)" "6#)*&\"\"$\"
\"\"\"\"%!\"\"*&F&F&F%F(" }{TEXT -1 72 ".\n\nThis is the same as the c
ube root of 3 divided by the cube root of 4." }}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 25 "denominator := 4 ^ (1/3);" }}}{PARA 0 "" 0 "" 
{TEXT -1 189 "\nWe want to rationalize the cube root of four. But four
 is already a square. So we have the cube root of two squared. If we h
ave two two's, how many more do we need to make three? Just one." }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "denominator * 2 ^ (1/3): % =
 simplify(%);" }}}{PARA 0 "" 0 "" {TEXT -1 78 "\nSo the method here is
 to multiply both top and bottom by the cube root of 2. " }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "(3/4)^(1/3);" }}}}{SECT 0 {PARA 4 "
" 0 "" {TEXT -1 30 "4. Denominators with Two Terms" }}{PARA 0 "" 0 "" 
{TEXT -1 368 "\nWhy do we need to rationalize radical expressions? Loo
k at these three different radical expressions. They all appear quite \+
different. However, look at the decimal values. Guess what... they are
 all equal! By having a consistent way of simplifying these expression
s, they will appear in the same form, and can be compared and seen to \+
be equal with much greater ease.\n" }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 133 "A:= (sqrt(2) + sqrt(6) )/ sqrt(3): % = evalf(%); \nB
:= sqrt(2/3) + sqrt(2) : % = evalf(%);\nC:= -4/(sqrt(6) - sqrt(18)) : \+
% = evalf(%);" }}}{PARA 0 "" 0 "" {TEXT -1 39 "\n\nHere are some typic
al expressioins :\n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "A := \+
1/(1 + sqrt(2));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "B := sq
rt(14)/(sqrt(7) - sqrt(2));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
26 "C := 1/(10 - 3*sqrt(29) );" }}}{PARA 0 "" 0 "" {TEXT -1 242 "\nHer
e is where we will use the difference of squares method we were lookin
g at before. We can rationalize any of these denomiators by multiplyin
g top and bottom by the CONJUGATE of the denominator. Let's examine th
is for each of these cases.\n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 217 "denom(A): `the denominator`  = %; \nconjugate_multiplier := op(
1,%%) - op(2,%%);\ndenom(A)*conjugate_multiplier  : % = expand(%);\nex
pand( numer(A)*conjugate_multiplier )\n / expand(denom(A)*conjugate_mu
ltiplier): A = %;\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 216 "den
om(B): `the denominator`  = %; \nconjugate_multiplier := op(1,%%) - op
(2,%%);\ndenom(B)*conjugate_multiplier  : % = expand(%);\nexpand( nume
r(B)*conjugate_multiplier )\n / expand(denom(B)*conjugate_multiplier):
 B = %;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 216 "denom(C): `the \+
denominator`  = %; \nconjugate_multiplier := op(1,%%) - op(2,%%);\nden
om(C)*conjugate_multiplier  : % = expand(%);\nexpand( numer(C)*conjuga
te_multiplier )\n / expand(denom(C)*conjugate_multiplier): C = %;" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 0 {PARA 4 "" 0 "
" {TEXT -1 43 "5. Denominators with Three Terms [optional]" }}{PARA 0 
"" 0 "" {TEXT -1 261 "\n[This is an optional section - not a normal pa
rt of the curriculum. However, with the power of Maple's algebraic tal
ents to ease the burden of algebraic computations, we can explore this
 topic that we might never choose to deal with in a traditional class.
]\n\n\n" }{TEXT 257 19 "A. Two Step Method\n" }{TEXT -1 108 " \nBreak \+
the three terms of the denominator into two groups - one with two term
s and the other with one term." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "Expr := 1/( 11 + 2*sqrt(7) -
 4*sqrt(3) );" }}}{PARA 0 "" 0 "" {TEXT -1 38 "\nWe can think of this \+
denominators as " }{XPPEDIT 18 0 "[11+2*sqrt(7)]-[4*sqrt(3)];" "6#,&7#
,&\"#6\"\"\"*&\"\"#F'-%%sqrtG6#\"\"(F'F'F'7#*&\"\"%F'-F+6#\"\"$F'!\"\"
" }{TEXT -1 33 ", and multiply by its conjugate, " }{XPPEDIT 18 0 "[11
+2*sqrt(7)]+[4*sqrt(3)];" "6#,&7#,&\"#6\"\"\"*&\"\"#F'-%%sqrtG6#\"\"(F
'F'F'7#*&\"\"%F'-F+6#\"\"$F'F'" }{TEXT -1 2 ".\n" }}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 41 "Multiplier := 11 + 2*sqrt(7) + 4*sqrt(3);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 96 "expand(numer(Expr)*Multiplie
r) / expand(denom(Expr)*Multiplier): Expr = %;\nS := simplify(%%): S;
" }}}{PARA 0 "" 0 "" {TEXT -1 74 "\nNow we multiply top and bottom by \+
the conjugate of this new denominator.\n" }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 31 "Multiplier := 101 - 44*sqrt(7);" }}}{PARA 0 "" 0 "" 
{TEXT -1 143 "\nThe result is an expression, while not pretty, at leas
t it is not tarnished by any irrational denominators. In other words, \+
we have succeeded!" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 78 "simpli
fy(\n    expand(numer(S)*Multiplier) / expand(denom(S)*Multiplier)    \+
 );" }}}{PARA 0 "" 0 "" {TEXT -1 44 "\nThere is also a way of doing it
 one step.\n\n" }{TEXT 256 1 "B" }{TEXT -1 2 ". " }{TEXT 258 27 "Ratio
nalizing Factor Method" }{TEXT -1 197 "\n\nWhen we rationalized a deno
minator with two terms, we multiplied top and bottom by a single expre
ssion. Is there such an expression for a three termed expression? Lets
 try the \"obvious\" choice..." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 39 "(a + b + c)*(a - b - c): % = expand(%);" }}}{PARA 0 "" 0 "" 
{TEXT -1 212 "\nIf b or c were radicals, then the resulting expression
 would still have radicals. The only way to eliminate all of the radic
als is to come up with an expression where ALL of the terms are square
d. Lets try this." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 77 "(a + b \+
+ c)*(a + b - c)*(a - b + c)*(a - b - c): % = sort(expand(%),[a,b,c]);
" }}}{PARA 0 "" 0 "" {TEXT -1 81 "\nThis works! All of the terms are s
quared or fourth powers. Lets try to use this." }}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 35 "Expr := 1/( 3 + sqrt(5) - sqrt(7));" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 132 "Multiplier := ( 3 + sqrt(5) + sqrt
(7)) * ( 3 - sqrt(5) + sqrt(7))\n                                     \+
  * ( 3 - sqrt(5) - sqrt(7));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 157 "`the rationalized denominator ` = expand( denom(Expr)*Multiplie
r );\nexpand( numer(Expr)*Multiplier) /  expand( denom(Expr)*Multiplie
r ): \nExpr = expand(%);  " }}}{PARA 0 "" 0 "" {TEXT -1 26 "\nHere is \+
another example :" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "Expr :=
 1/( 14*sqrt(2) - 8*sqrt(11) + 3*sqrt(17));" }}}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 166 "Multiplier :=   (14*sqrt(2) + 8*sqrt(11) + 3*sqrt
(17))\n               *(14*sqrt(2) + 8*sqrt(11) - 3*sqrt(17))\n       \+
        *(14*sqrt(2) - 8*sqrt(11) - 3*sqrt(17)) ;" }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 155 "`the rationalized denominator ` = expand( d
enom(Expr)*Multiplier );\nexpand( numer(Expr)*Multiplier) /  expand( d
enom(Expr)*Multiplier ): \nExpr = expand(%);" }}}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}}{PARA 0 "" 0 "" {TEXT 265 36 "\n         \251 2002 Wa
terloo Maple Inc " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{MARK "0 0" 31 }
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